m38 lec 021814
TRANSCRIPT
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MATH 38: Mathematical Analysis IIIUnit 3: Differentiation of Functions of More than One Variable
I. F. Evidente
IMSP (UPLB)
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Outline
1 Relative Extreme Function Values
2 Absolute Extrema of Functions of Two Variables
Figures taken from: J. Stewart, The Calculus: Early Transcendentals,Brooks/Cole, 6th Edition, 2008.
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Outline
1 Relative Extreme Function Values
2 Absolute Extrema of Functions of Two Variables
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DefinitionGiven a function of two variables f (x, y) and a point (x0, y0) on thexy-plane:
1 f is said to have a relative maximum at (x0, y0) if
there exists anopen ball B on the xy-plane centered at (x0, y0) such thatf (x0, y0) f (x, y) for every (x, y) B .
2 f is said to have a relative minumum at (x0, y0) if there exists anopen ball B on the xy-plane centered at (x0, y0) such thatf (x0, y0) f (x, y) for every (x, y) B
3 f is said to have a relative extremum at (x0, y0) if either f has arelative maximum or a relative minimum at (x0, y0).
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DefinitionGiven a function of two variables f (x, y) and a point (x0, y0) on thexy-plane:
1 f is said to have a relative maximum at (x0, y0) if there exists anopen ball B on the xy-plane centered at (x0, y0) such that
f (x0, y0) f (x, y) for every (x, y) B .2 f is said to have a relative minumum at (x0, y0) if there exists an
open ball B on the xy-plane centered at (x0, y0) such thatf (x0, y0) f (x, y) for every (x, y) B
3 f is said to have a relative extremum at (x0, y0) if either f has arelative maximum or a relative minimum at (x0, y0).
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DefinitionGiven a function of two variables f (x, y) and a point (x0, y0) on thexy-plane:
1 f is said to have a relative maximum at (x0, y0) if there exists anopen ball B on the xy-plane centered at (x0, y0) such thatf (x0, y0) f (x, y) for every (x, y) B .
2 f is said to have a relative minumum at (x0, y0) if there exists anopen ball B on the xy-plane centered at (x0, y0) such thatf (x0, y0) f (x, y) for every (x, y) B
3 f is said to have a relative extremum at (x0, y0) if either f has arelative maximum or a relative minimum at (x0, y0).
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DefinitionGiven a function of two variables f (x, y) and a point (x0, y0) on thexy-plane:
1 f is said to have a relative maximum at (x0, y0) if there exists anopen ball B on the xy-plane centered at (x0, y0) such thatf (x0, y0) f (x, y) for every (x, y) B .
2 f is said to have a relative minumum at (x0, y0) if
there exists anopen ball B on the xy-plane centered at (x0, y0) such thatf (x0, y0) f (x, y) for every (x, y) B
3 f is said to have a relative extremum at (x0, y0) if either f has arelative maximum or a relative minimum at (x0, y0).
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DefinitionGiven a function of two variables f (x, y) and a point (x0, y0) on thexy-plane:
1 f is said to have a relative maximum at (x0, y0) if there exists anopen ball B on the xy-plane centered at (x0, y0) such thatf (x0, y0) f (x, y) for every (x, y) B .
2 f is said to have a relative minumum at (x0, y0) if there exists anopen ball B on the xy-plane centered at (x0, y0) such that
f (x0, y0) f (x, y) for every (x, y) B3 f is said to have a relative extremum at (x0, y0) if either f has a
relative maximum or a relative minimum at (x0, y0).
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DefinitionGiven a function of two variables f (x, y) and a point (x0, y0) on thexy-plane:
1 f is said to have a relative maximum at (x0, y0) if there exists anopen ball B on the xy-plane centered at (x0, y0) such thatf (x0, y0) f (x, y) for every (x, y) B .
2 f is said to have a relative minumum at (x0, y0) if there exists anopen ball B on the xy-plane centered at (x0, y0) such thatf (x0, y0) f (x, y) for every (x, y) B
3 f is said to have a relative extremum at (x0, y0) if either f has arelative maximum or a relative minimum at (x0, y0).
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DefinitionGiven a function of two variables f (x, y) and a point (x0, y0) on thexy-plane:
1 f is said to have a relative maximum at (x0, y0) if there exists anopen ball B on the xy-plane centered at (x0, y0) such thatf (x0, y0) f (x, y) for every (x, y) B .
2 f is said to have a relative minumum at (x0, y0) if there exists anopen ball B on the xy-plane centered at (x0, y0) such thatf (x0, y0) f (x, y) for every (x, y) B
3 f is said to have a relative extremum at (x0, y0) if
either f has arelative maximum or a relative minimum at (x0, y0).
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DefinitionGiven a function of two variables f (x, y) and a point (x0, y0) on thexy-plane:
1 f is said to have a relative maximum at (x0, y0) if there exists anopen ball B on the xy-plane centered at (x0, y0) such thatf (x0, y0) f (x, y) for every (x, y) B .
2 f is said to have a relative minumum at (x0, y0) if there exists anopen ball B on the xy-plane centered at (x0, y0) such thatf (x0, y0) f (x, y) for every (x, y) B
3 f is said to have a relative extremum at (x0, y0) if either f has arelative maximum or a relative minimum at (x0, y0).
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RemarkSuppose f has a relative minimum [maximum, extremum] at P .
The value f (x0, y0) is said to be a relative minimum [maximum,extremum] value of f .The point (x0, y0, f (x0, y0)) is called a relative minimum [maximum,extremum] point of f .
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RemarkSuppose f has a relative minimum [maximum, extremum] at P .
The value f (x0, y0) is said to be a relative minimum [maximum,extremum] value of f .
The point (x0, y0, f (x0, y0)) is called a relative minimum [maximum,extremum] point of f .
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RemarkSuppose f has a relative minimum [maximum, extremum] at P .
The value f (x0, y0) is said to be a relative minimum [maximum,extremum] value of f .The point (x0, y0, f (x0, y0)) is called a relative minimum [maximum,extremum] point of f .
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Graphically, the graph of z = f (x, y) has peaks and trenches at relativemaximum and relative minimum points, respectively.
It is "relative" or "local" because the point is highest or lowest relative toits very close neighbors.Note that a function may have several relative extrema.
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Graphically, the graph of z = f (x, y) has peaks and trenches at relativemaximum and relative minimum points, respectively.
It is "relative" or "local" because the point is highest or lowest relative toits very close neighbors.
Note that a function may have several relative extrema.
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Graphically, the graph of z = f (x, y) has peaks and trenches at relativemaximum and relative minimum points, respectively.
It is "relative" or "local" because the point is highest or lowest relative toits very close neighbors.Note that a function may have several relative extrema.
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Problem:How do we find the relative minima and relative maxima of f ?
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DefinitionThe point (x0, y0) is a critical point of the function f (x, y) if
(x0, y0) dom f and either1 fx(x0, y0)= 0 and fy (x0, y0)= 0 (Type 1), or2 fx(x0, y0) or fy (x0, y0) does not exist (Type 2)
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DefinitionThe point (x0, y0) is a critical point of the function f (x, y) if(x0, y0) dom f and either
1 fx(x0, y0)= 0 and fy (x0, y0)= 0 (Type 1), or2 fx(x0, y0) or fy (x0, y0) does not exist (Type 2)
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DefinitionThe point (x0, y0) is a critical point of the function f (x, y) if(x0, y0) dom f and either
1 fx(x0, y0)= 0 and fy (x0, y0)= 0 (Type 1), or
2 fx(x0, y0) or fy (x0, y0) does not exist (Type 2)
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DefinitionThe point (x0, y0) is a critical point of the function f (x, y) if(x0, y0) dom f and either
1 fx(x0, y0)= 0 and fy (x0, y0)= 0 (Type 1), or2 fx(x0, y0) or fy (x0, y0) does not exist (Type 2)
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ExampleFind the critical points of f (x, y)= 3x22xy + y28y .
fx(x, y)= 6x2y and fy (x, y)= 2x+2y 8Critical Points:
6x2y = 02x+2y 8 = 0
} 4x8= 0 x = 2 & y = 6
(2,6) is a critical point of f .
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ExampleFind the critical points of f (x, y)= 3x22xy + y28y .
fx(x, y)= 6x
2y and fy (x, y)= 2x+2y 8Critical Points:
6x2y = 02x+2y 8 = 0
} 4x8= 0 x = 2 & y = 6
(2,6) is a critical point of f .
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ExampleFind the critical points of f (x, y)= 3x22xy + y28y .
fx(x, y)= 6x2y
and fy (x, y)= 2x+2y 8Critical Points:
6x2y = 02x+2y 8 = 0
} 4x8= 0 x = 2 & y = 6
(2,6) is a critical point of f .
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ExampleFind the critical points of f (x, y)= 3x22xy + y28y .
fx(x, y)= 6x2y and fy (x, y)=
2x+2y 8Critical Points:
6x2y = 02x+2y 8 = 0
} 4x8= 0 x = 2 & y = 6
(2,6) is a critical point of f .
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ExampleFind the critical points of f (x, y)= 3x22xy + y28y .
fx(x, y)= 6x2y and fy (x, y)= 2x
+2y 8Critical Points:
6x2y = 02x+2y 8 = 0
} 4x8= 0 x = 2 & y = 6
(2,6) is a critical point of f .
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ExampleFind the critical points of f (x, y)= 3x22xy + y28y .
fx(x, y)= 6x2y and fy (x, y)= 2x+2y
8Critical Points:
6x2y = 02x+2y 8 = 0
} 4x8= 0 x = 2 & y = 6
(2,6) is a critical point of f .
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ExampleFind the critical points of f (x, y)= 3x22xy + y28y .
fx(x, y)= 6x2y and fy (x, y)= 2x+2y 8
Critical Points:
6x2y = 02x+2y 8 = 0
} 4x8= 0 x = 2 & y = 6
(2,6) is a critical point of f .
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ExampleFind the critical points of f (x, y)= 3x22xy + y28y .
fx(x, y)= 6x2y and fy (x, y)= 2x+2y 8Critical Points:
6x2y = 02x+2y 8 = 0
} 4x8= 0 x = 2 & y = 6
(2,6) is a critical point of f .
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ExampleFind the critical points of f (x, y)= 3x22xy + y28y .
fx(x, y)= 6x2y and fy (x, y)= 2x+2y 8Critical Points:
6x2y = 02x+2y 8 = 0
}
4x8= 0 x = 2 & y = 6
(2,6) is a critical point of f .
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ExampleFind the critical points of f (x, y)= 3x22xy + y28y .
fx(x, y)= 6x2y and fy (x, y)= 2x+2y 8Critical Points:
6x2y = 02x+2y 8 = 0
} 4x8= 0
x = 2 & y = 6
(2,6) is a critical point of f .
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ExampleFind the critical points of f (x, y)= 3x22xy + y28y .
fx(x, y)= 6x2y and fy (x, y)= 2x+2y 8Critical Points:
6x2y = 02x+2y 8 = 0
} 4x8= 0 x = 2 &
y = 6
(2,6) is a critical point of f .
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ExampleFind the critical points of f (x, y)= 3x22xy + y28y .
fx(x, y)= 6x2y and fy (x, y)= 2x+2y 8Critical Points:
6x2y = 02x+2y 8 = 0
} 4x8= 0 x = 2 & y = 6
(2,6) is a critical point of f .
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ExampleFind the critical points of f (x, y)= 3x22xy + y28y .
fx(x, y)= 6x2y and fy (x, y)= 2x+2y 8Critical Points:
6x2y = 02x+2y 8 = 0
} 4x8= 0 x = 2 & y = 6
(2,6) is a critical point of f .
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TheoremIf the function f (x, y) has a relative extremum at (x0,y0) and fx(x0, y0) andfy (x0, y0) exist, then
fx(x0, y0)= 0 and fy (x0, y0)= 0.
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TheoremIf the function f (x, y) has a relative extremum at (x0,y0) and fx(x0, y0) andfy (x0, y0) exist, then fx(x0, y0)= 0 and fy (x0, y0)= 0.
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The theorem tells us the following:
RemarkIf f (x, y) has a relative extrema at (x0, y0) and the partial derivativesof f exist at that point, then (x0, y0) must be a Type 1 critical point.If f is differentiable for all points in its domain, then the candidatesfor points where relative extrema occur come from the critical pointsof f . (Naturally, these are all Type 1 critical points) Venn Diagram?The converse of the theorem is NOT true. It is not true that if (x0, y0)is a critical point of f , then f has a relative extrema at (x0, y0).
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The theorem tells us the following:
RemarkIf f (x, y) has a relative extrema at (x0, y0) and the partial derivativesof f exist at that point, then (x0, y0) must be
a Type 1 critical point.If f is differentiable for all points in its domain, then the candidatesfor points where relative extrema occur come from the critical pointsof f . (Naturally, these are all Type 1 critical points) Venn Diagram?The converse of the theorem is NOT true. It is not true that if (x0, y0)is a critical point of f , then f has a relative extrema at (x0, y0).
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The theorem tells us the following:
RemarkIf f (x, y) has a relative extrema at (x0, y0) and the partial derivativesof f exist at that point, then (x0, y0) must be a Type 1 critical point.
If f is differentiable for all points in its domain, then the candidatesfor points where relative extrema occur come from the critical pointsof f . (Naturally, these are all Type 1 critical points) Venn Diagram?The converse of the theorem is NOT true. It is not true that if (x0, y0)is a critical point of f , then f has a relative extrema at (x0, y0).
-
The theorem tells us the following:
RemarkIf f (x, y) has a relative extrema at (x0, y0) and the partial derivativesof f exist at that point, then (x0, y0) must be a Type 1 critical point.If f is differentiable for all points in its domain, then the candidatesfor points where relative extrema occur come from
the critical pointsof f . (Naturally, these are all Type 1 critical points) Venn Diagram?The converse of the theorem is NOT true. It is not true that if (x0, y0)is a critical point of f , then f has a relative extrema at (x0, y0).
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The theorem tells us the following:
RemarkIf f (x, y) has a relative extrema at (x0, y0) and the partial derivativesof f exist at that point, then (x0, y0) must be a Type 1 critical point.If f is differentiable for all points in its domain, then the candidatesfor points where relative extrema occur come from the critical pointsof f .
(Naturally, these are all Type 1 critical points) Venn Diagram?The converse of the theorem is NOT true. It is not true that if (x0, y0)is a critical point of f , then f has a relative extrema at (x0, y0).
-
The theorem tells us the following:
RemarkIf f (x, y) has a relative extrema at (x0, y0) and the partial derivativesof f exist at that point, then (x0, y0) must be a Type 1 critical point.If f is differentiable for all points in its domain, then the candidatesfor points where relative extrema occur come from the critical pointsof f . (Naturally, these are all Type 1 critical points)
Venn Diagram?The converse of the theorem is NOT true. It is not true that if (x0, y0)is a critical point of f , then f has a relative extrema at (x0, y0).
-
The theorem tells us the following:
RemarkIf f (x, y) has a relative extrema at (x0, y0) and the partial derivativesof f exist at that point, then (x0, y0) must be a Type 1 critical point.If f is differentiable for all points in its domain, then the candidatesfor points where relative extrema occur come from the critical pointsof f . (Naturally, these are all Type 1 critical points) Venn Diagram?
The converse of the theorem is NOT true. It is not true that if (x0, y0)is a critical point of f , then f has a relative extrema at (x0, y0).
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The theorem tells us the following:
RemarkIf f (x, y) has a relative extrema at (x0, y0) and the partial derivativesof f exist at that point, then (x0, y0) must be a Type 1 critical point.If f is differentiable for all points in its domain, then the candidatesfor points where relative extrema occur come from the critical pointsof f . (Naturally, these are all Type 1 critical points) Venn Diagram?The converse of the theorem is NOT true. It is not true that if (x0, y0)is a critical point of f , then f has a relative extrema at (x0, y0).
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Geometrically
Suppose the following:z = f (x, y) has a relative extremum at (x0, y0)The tangent plane of the graph of f at this point exists.
fx(x0, y0)= 0 and fy (x0, y0)= 0Equation of tangent plane:
z = L(x, y)z = fx(x0, y0)(xx0)+ fy (x0, y0)(y y0)+ f (x0, y0)z = f (x0, y0)
What kind of plane is this? HORIZONTAL!
RemarkIf the tangent plane of the graph of f at a relative extreme point exists,then it must be horizontal!
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Geometrically
Suppose the following:z = f (x, y) has a relative extremum at (x0, y0)The tangent plane of the graph of f at this point exists.
fx(x0, y0)= 0 and fy (x0, y0)= 0
Equation of tangent plane:
z = L(x, y)z = fx(x0, y0)(xx0)+ fy (x0, y0)(y y0)+ f (x0, y0)z = f (x0, y0)
What kind of plane is this? HORIZONTAL!
RemarkIf the tangent plane of the graph of f at a relative extreme point exists,then it must be horizontal!
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Geometrically
Suppose the following:z = f (x, y) has a relative extremum at (x0, y0)The tangent plane of the graph of f at this point exists.
fx(x0, y0)= 0 and fy (x0, y0)= 0Equation of tangent plane:
z = L(x, y)
z = fx(x0, y0)(xx0)+ fy (x0, y0)(y y0)+ f (x0, y0)z = f (x0, y0)
What kind of plane is this? HORIZONTAL!
RemarkIf the tangent plane of the graph of f at a relative extreme point exists,then it must be horizontal!
-
Geometrically
Suppose the following:z = f (x, y) has a relative extremum at (x0, y0)The tangent plane of the graph of f at this point exists.
fx(x0, y0)= 0 and fy (x0, y0)= 0Equation of tangent plane:
z = L(x, y)z = fx(x0, y0)(xx0)+ fy (x0, y0)(y y0)+ f (x0, y0)
z = f (x0, y0)
What kind of plane is this? HORIZONTAL!
RemarkIf the tangent plane of the graph of f at a relative extreme point exists,then it must be horizontal!
-
Geometrically
Suppose the following:z = f (x, y) has a relative extremum at (x0, y0)The tangent plane of the graph of f at this point exists.
fx(x0, y0)= 0 and fy (x0, y0)= 0Equation of tangent plane:
z = L(x, y)z = fx(x0, y0)(xx0)+ fy (x0, y0)(y y0)+ f (x0, y0)z = f (x0, y0)
What kind of plane is this? HORIZONTAL!
RemarkIf the tangent plane of the graph of f at a relative extreme point exists,then it must be horizontal!
-
Geometrically
Suppose the following:z = f (x, y) has a relative extremum at (x0, y0)The tangent plane of the graph of f at this point exists.
fx(x0, y0)= 0 and fy (x0, y0)= 0Equation of tangent plane:
z = L(x, y)z = fx(x0, y0)(xx0)+ fy (x0, y0)(y y0)+ f (x0, y0)z = f (x0, y0)
What kind of plane is this?
HORIZONTAL!
RemarkIf the tangent plane of the graph of f at a relative extreme point exists,then it must be horizontal!
-
Geometrically
Suppose the following:z = f (x, y) has a relative extremum at (x0, y0)The tangent plane of the graph of f at this point exists.
fx(x0, y0)= 0 and fy (x0, y0)= 0Equation of tangent plane:
z = L(x, y)z = fx(x0, y0)(xx0)+ fy (x0, y0)(y y0)+ f (x0, y0)z = f (x0, y0)
What kind of plane is this? HORIZONTAL!
RemarkIf the tangent plane of the graph of f at a relative extreme point exists,then it must be horizontal!
-
Geometrically
Suppose the following:z = f (x, y) has a relative extremum at (x0, y0)The tangent plane of the graph of f at this point exists.
fx(x0, y0)= 0 and fy (x0, y0)= 0Equation of tangent plane:
z = L(x, y)z = fx(x0, y0)(xx0)+ fy (x0, y0)(y y0)+ f (x0, y0)z = f (x0, y0)
What kind of plane is this? HORIZONTAL!
RemarkIf the tangent plane of the graph of f at a relative extreme point exists,then
it must be horizontal!
-
Geometrically
Suppose the following:z = f (x, y) has a relative extremum at (x0, y0)The tangent plane of the graph of f at this point exists.
fx(x0, y0)= 0 and fy (x0, y0)= 0Equation of tangent plane:
z = L(x, y)z = fx(x0, y0)(xx0)+ fy (x0, y0)(y y0)+ f (x0, y0)z = f (x0, y0)
What kind of plane is this? HORIZONTAL!
RemarkIf the tangent plane of the graph of f at a relative extreme point exists,then it must be horizontal!
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DefinitionA point (x0, y0,z0) is said to be a saddle point of a surface if
there existtwo vertical planes such that the trace of in one plane has a relativemaximum at (x0, y0) and the trace of in the other plane has a relativeminimum at (x0, y0).
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DefinitionA point (x0, y0,z0) is said to be a saddle point of a surface if there existtwo vertical planes such that
the trace of in one plane has a relativemaximum at (x0, y0) and the trace of in the other plane has a relativeminimum at (x0, y0).
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DefinitionA point (x0, y0,z0) is said to be a saddle point of a surface if there existtwo vertical planes such that the trace of in one plane has a relativemaximum at (x0, y0) and
the trace of in the other plane has a relativeminimum at (x0, y0).
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DefinitionA point (x0, y0,z0) is said to be a saddle point of a surface if there existtwo vertical planes such that the trace of in one plane has a relativemaximum at (x0, y0) and the trace of in the other plane has a relativeminimum at (x0, y0).
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f (x, y)= x2 y2 has a saddle point at (0,0)
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Theorem (Second Derivative Test for Relative Extrema)Suppose that the first and second order partial derivatives of adifferentiable function f (x, y) is continuous on some open ball B
((x0, y0),r
)and fx(x0, y0)= fy (x0, y0)= 0.
Let
D = fxx(x0, y0) fy y (x0, y0)[fxy (x0, y0)
]2 .1 If D > 0 and
1 fxx (x0, y0)> 0, then f (x, y) has a relative minimum at (x0, y0).2 fxx (x0, y0)< 0, then f (x, y) has a relative maximum at (x0, y0).
2 If D < 0, then f (x, y) has a saddle point at (x0, y0).3 If D = 0, then no conclusion can be made.
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Theorem (Second Derivative Test for Relative Extrema)Suppose that the first and second order partial derivatives of adifferentiable function f (x, y) is continuous on some open ball B
((x0, y0),r
)and fx(x0, y0)= fy (x0, y0)= 0. Let
D = fxx(x0, y0) fy y (x0, y0)[fxy (x0, y0)
]2 .
1 If D > 0 and1 fxx (x0, y0)> 0, then f (x, y) has a relative minimum at (x0, y0).2 fxx (x0, y0)< 0, then f (x, y) has a relative maximum at (x0, y0).
2 If D < 0, then f (x, y) has a saddle point at (x0, y0).3 If D = 0, then no conclusion can be made.
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Theorem (Second Derivative Test for Relative Extrema)Suppose that the first and second order partial derivatives of adifferentiable function f (x, y) is continuous on some open ball B
((x0, y0),r
)and fx(x0, y0)= fy (x0, y0)= 0. Let
D = fxx(x0, y0) fy y (x0, y0)[fxy (x0, y0)
]2 .1 If D > 0 and
1 fxx (x0, y0)> 0, then f (x, y) has a relative minimum at (x0, y0).2 fxx (x0, y0)< 0, then f (x, y) has a relative maximum at (x0, y0).
2 If D < 0, then f (x, y) has a saddle point at (x0, y0).3 If D = 0, then no conclusion can be made.
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Theorem (Second Derivative Test for Relative Extrema)Suppose that the first and second order partial derivatives of adifferentiable function f (x, y) is continuous on some open ball B
((x0, y0),r
)and fx(x0, y0)= fy (x0, y0)= 0. Let
D = fxx(x0, y0) fy y (x0, y0)[fxy (x0, y0)
]2 .1 If D > 0 and
1 fxx (x0, y0)> 0, then
f (x, y) has a relative minimum at (x0, y0).2 fxx (x0, y0)< 0, then f (x, y) has a relative maximum at (x0, y0).
2 If D < 0, then f (x, y) has a saddle point at (x0, y0).3 If D = 0, then no conclusion can be made.
-
Theorem (Second Derivative Test for Relative Extrema)Suppose that the first and second order partial derivatives of adifferentiable function f (x, y) is continuous on some open ball B
((x0, y0),r
)and fx(x0, y0)= fy (x0, y0)= 0. Let
D = fxx(x0, y0) fy y (x0, y0)[fxy (x0, y0)
]2 .1 If D > 0 and
1 fxx (x0, y0)> 0, then f (x, y) has a relative minimum at (x0, y0).
2 fxx (x0, y0)< 0, then f (x, y) has a relative maximum at (x0, y0).2 If D < 0, then f (x, y) has a saddle point at (x0, y0).3 If D = 0, then no conclusion can be made.
-
Theorem (Second Derivative Test for Relative Extrema)Suppose that the first and second order partial derivatives of adifferentiable function f (x, y) is continuous on some open ball B
((x0, y0),r
)and fx(x0, y0)= fy (x0, y0)= 0. Let
D = fxx(x0, y0) fy y (x0, y0)[fxy (x0, y0)
]2 .1 If D > 0 and
1 fxx (x0, y0)> 0, then f (x, y) has a relative minimum at (x0, y0).2 fxx (x0, y0)< 0, then
f (x, y) has a relative maximum at (x0, y0).2 If D < 0, then f (x, y) has a saddle point at (x0, y0).3 If D = 0, then no conclusion can be made.
-
Theorem (Second Derivative Test for Relative Extrema)Suppose that the first and second order partial derivatives of adifferentiable function f (x, y) is continuous on some open ball B
((x0, y0),r
)and fx(x0, y0)= fy (x0, y0)= 0. Let
D = fxx(x0, y0) fy y (x0, y0)[fxy (x0, y0)
]2 .1 If D > 0 and
1 fxx (x0, y0)> 0, then f (x, y) has a relative minimum at (x0, y0).2 fxx (x0, y0)< 0, then f (x, y) has a relative maximum at (x0, y0).
2 If D < 0, then f (x, y) has a saddle point at (x0, y0).3 If D = 0, then no conclusion can be made.
-
Theorem (Second Derivative Test for Relative Extrema)Suppose that the first and second order partial derivatives of adifferentiable function f (x, y) is continuous on some open ball B
((x0, y0),r
)and fx(x0, y0)= fy (x0, y0)= 0. Let
D = fxx(x0, y0) fy y (x0, y0)[fxy (x0, y0)
]2 .1 If D > 0 and
1 fxx (x0, y0)> 0, then f (x, y) has a relative minimum at (x0, y0).2 fxx (x0, y0)< 0, then f (x, y) has a relative maximum at (x0, y0).
2 If D < 0, then
f (x, y) has a saddle point at (x0, y0).3 If D = 0, then no conclusion can be made.
-
Theorem (Second Derivative Test for Relative Extrema)Suppose that the first and second order partial derivatives of adifferentiable function f (x, y) is continuous on some open ball B
((x0, y0),r
)and fx(x0, y0)= fy (x0, y0)= 0. Let
D = fxx(x0, y0) fy y (x0, y0)[fxy (x0, y0)
]2 .1 If D > 0 and
1 fxx (x0, y0)> 0, then f (x, y) has a relative minimum at (x0, y0).2 fxx (x0, y0)< 0, then f (x, y) has a relative maximum at (x0, y0).
2 If D < 0, then f (x, y) has a saddle point at (x0, y0).
3 If D = 0, then no conclusion can be made.
-
Theorem (Second Derivative Test for Relative Extrema)Suppose that the first and second order partial derivatives of adifferentiable function f (x, y) is continuous on some open ball B
((x0, y0),r
)and fx(x0, y0)= fy (x0, y0)= 0. Let
D = fxx(x0, y0) fy y (x0, y0)[fxy (x0, y0)
]2 .1 If D > 0 and
1 fxx (x0, y0)> 0, then f (x, y) has a relative minimum at (x0, y0).2 fxx (x0, y0)< 0, then f (x, y) has a relative maximum at (x0, y0).
2 If D < 0, then f (x, y) has a saddle point at (x0, y0).3 If D = 0, then
no conclusion can be made.
-
Theorem (Second Derivative Test for Relative Extrema)Suppose that the first and second order partial derivatives of adifferentiable function f (x, y) is continuous on some open ball B
((x0, y0),r
)and fx(x0, y0)= fy (x0, y0)= 0. Let
D = fxx(x0, y0) fy y (x0, y0)[fxy (x0, y0)
]2 .1 If D > 0 and
1 fxx (x0, y0)> 0, then f (x, y) has a relative minimum at (x0, y0).2 fxx (x0, y0)< 0, then f (x, y) has a relative maximum at (x0, y0).
2 If D < 0, then f (x, y) has a saddle point at (x0, y0).3 If D = 0, then no conclusion can be made.
-
RemarkThis test is only for Type 1 critical points!If f is differentiable, then this test allows you to get ALL criticalpoints.
-
RemarkOne way to remember the formula for D:
The Jacobian of f is the matrix of partial derivatives of f[fxx fxyfyx fy y
]D is the determinant of the Jacobian of f .
TheoremLet f be a function in 2 variables and R a region on the xy-plane. If fxyand fyx are continuous on R, then fxy = fyx on that open ball.
-
RemarkOne way to remember the formula for D:The Jacobian of f is the matrix of partial derivatives of f[
fxx fxyfyx fy y
]
D is the determinant of the Jacobian of f .
TheoremLet f be a function in 2 variables and R a region on the xy-plane. If fxyand fyx are continuous on R, then fxy = fyx on that open ball.
-
RemarkOne way to remember the formula for D:The Jacobian of f is the matrix of partial derivatives of f[
fxx fxyfyx fy y
]D is the determinant of the Jacobian of f .
TheoremLet f be a function in 2 variables and R a region on the xy-plane. If fxyand fyx are continuous on R, then fxy = fyx on that open ball.
-
RemarkOne way to remember the formula for D:The Jacobian of f is the matrix of partial derivatives of f[
fxx fxyfyx fy y
]D is the determinant of the Jacobian of f .
TheoremLet f be a function in 2 variables and R a region on the xy-plane. If fxyand fyx are continuous on R, then fxy = fyx on that open ball.
-
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)= 2Thus,
D = 6 22 2
= 124= 8> 0 (rel ext)Now
fxx(2,6)= 6> 0 f has a relative minimum at (2,6)
-
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8
Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)= 2Thus,
D = 6 22 2
= 124= 8> 0 (rel ext)Now
fxx(2,6)= 6> 0 f has a relative minimum at (2,6)
-
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point:
(2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)= 2Thus,
D = 6 22 2
= 124= 8> 0 (rel ext)Now
fxx(2,6)= 6> 0 f has a relative minimum at (2,6)
-
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)
Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)= 2Thus,
D = 6 22 2
= 124= 8> 0 (rel ext)Now
fxx(2,6)= 6> 0 f has a relative minimum at (2,6)
-
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)=
6 fy y (x, y)= 2 fxy (x, y)= 2Thus,
D = 6 22 2
= 124= 8> 0 (rel ext)Now
fxx(2,6)= 6> 0 f has a relative minimum at (2,6)
-
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6
fy y (x, y)= 2 fxy (x, y)= 2Thus,
D = 6 22 2
= 124= 8> 0 (rel ext)Now
fxx(2,6)= 6> 0 f has a relative minimum at (2,6)
-
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)=
2 fxy (x, y)= 2Thus,
D = 6 22 2
= 124= 8> 0 (rel ext)Now
fxx(2,6)= 6> 0 f has a relative minimum at (2,6)
-
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2
fxy (x, y)= 2Thus,
D = 6 22 2
= 124= 8> 0 (rel ext)Now
fxx(2,6)= 6> 0 f has a relative minimum at (2,6)
-
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)=
2Thus,
D = 6 22 2
= 124= 8> 0 (rel ext)Now
fxx(2,6)= 6> 0 f has a relative minimum at (2,6)
-
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)= 2
Thus,
D = 6 22 2
= 124= 8> 0 (rel ext)Now
fxx(2,6)= 6> 0 f has a relative minimum at (2,6)
-
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)= 2Thus,
D = 6 22 2
= 124= 8> 0 (rel ext)Now
fxx(2,6)= 6> 0 f has a relative minimum at (2,6)
-
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)= 2Thus,
D = 6
22 2
= 124= 8> 0 (rel ext)Now
fxx(2,6)= 6> 0 f has a relative minimum at (2,6)
-
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)= 2Thus,
D = 6
22 2
= 124= 8> 0 (rel ext)Now
fxx(2,6)= 6> 0 f has a relative minimum at (2,6)
-
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)= 2Thus,
D = 6 2
2 2= 124= 8> 0 (rel ext)
Nowfxx(2,6)= 6> 0
f has a relative minimum at (2,6)
-
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)= 2Thus,
D = 6 22
2
= 124= 8> 0 (rel ext)Now
fxx(2,6)= 6> 0 f has a relative minimum at (2,6)
-
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)= 2Thus,
D = 6 22 2
=
124= 8> 0 (rel ext)
Nowfxx(2,6)= 6> 0
f has a relative minimum at (2,6)
-
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)= 2Thus,
D = 6 22 2
= 12
4= 8> 0 (rel ext)
Nowfxx(2,6)= 6> 0
f has a relative minimum at (2,6)
-
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)= 2Thus,
D = 6 22 2
= 124=
8> 0 (rel ext)
Nowfxx(2,6)= 6> 0
f has a relative minimum at (2,6)
-
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)= 2Thus,
D = 6 22 2
= 124= 8
> 0 (rel ext)
Nowfxx(2,6)= 6> 0
f has a relative minimum at (2,6)
-
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)= 2Thus,
D = 6 22 2
= 124= 8> 0
(rel ext)
Nowfxx(2,6)= 6> 0
f has a relative minimum at (2,6)
-
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)= 2Thus,
D = 6 22 2
= 124= 8> 0 (rel ext)
Nowfxx(2,6)= 6> 0
f has a relative minimum at (2,6)
-
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)= 2Thus,
D = 6 22 2
= 124= 8> 0 (rel ext)Now
fxx(2,6)=
6> 0 f has a relative minimum at (2,6)
-
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)= 2Thus,
D = 6 22 2
= 124= 8> 0 (rel ext)Now
fxx(2,6)= 6
> 0 f has a relative minimum at (2,6)
-
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)= 2Thus,
D = 6 22 2
= 124= 8> 0 (rel ext)Now
fxx(2,6)= 6> 0
f has a relative minimum at (2,6)
-
ExampleLocate all relative extrema and saddle points of
f (x, y)= 3x22xy + y28y
.
fx(x, y)= 6x2y and fy (x, y)=2x+2y 8Critical Point: (2,6)Second Partial Derivatives:
fxx(x, y)= 6 fy y (x, y)= 2 fxy (x, y)= 2Thus,
D = 6 22 2
= 124= 8> 0 (rel ext)Now
fxx(2,6)= 6> 0 f has a relative minimum at (2,6)
-
ExampleLocate all relative extrema and saddle points of f (x, y)= 4xy x4 y4.
fx(x, y)= 4y 4x3 and fy (x, y)= 4x4y3
Critical Points:
4y 4x3 = 0 (1)4x4y3 = 0 (2)
} y = x
3 (1)x = y3 (2)
}
-
ExampleLocate all relative extrema and saddle points of f (x, y)= 4xy x4 y4.
fx(x, y)=
4y 4x3 and fy (x, y)= 4x4y3
Critical Points:
4y 4x3 = 0 (1)4x4y3 = 0 (2)
} y = x
3 (1)x = y3 (2)
}
-
ExampleLocate all relative extrema and saddle points of f (x, y)= 4xy x4 y4.
fx(x, y)= 4y 4x3
and fy (x, y)= 4x4y3
Critical Points:
4y 4x3 = 0 (1)4x4y3 = 0 (2)
} y = x
3 (1)x = y3 (2)
}
-
ExampleLocate all relative extrema and saddle points of f (x, y)= 4xy x4 y4.
fx(x, y)= 4y 4x3 and
fy (x, y)= 4x4y3
Critical Points:
4y 4x3 = 0 (1)4x4y3 = 0 (2)
} y = x
3 (1)x = y3 (2)
}
-
ExampleLocate all relative extrema and saddle points of f (x, y)= 4xy x4 y4.
fx(x, y)= 4y 4x3 and fy (x, y)=
4x4y3
Critical Points:
4y 4x3 = 0 (1)4x4y3 = 0 (2)
} y = x
3 (1)x = y3 (2)
}
-
ExampleLocate all relative extrema and saddle points of f (x, y)= 4xy x4 y4.
fx(x, y)= 4y 4x3 and fy (x, y)= 4x4y3
Critical Points:
4y 4x3 = 0 (1)4x4y3 = 0 (2)
} y = x
3 (1)x = y3 (2)
}
-
ExampleLocate all relative extrema and saddle points of f (x, y)= 4xy x4 y4.
fx(x, y)= 4y 4x3 and fy (x, y)= 4x4y3
Critical Points:
4y 4x3 = 0 (1)4x4y3 = 0 (2)
} y = x
3 (1)x = y3 (2)
}
-
ExampleLocate all relative extrema and saddle points of f (x, y)= 4xy x4 y4.
fx(x, y)= 4y 4x3 and fy (x, y)= 4x4y3
Critical Points:
4y 4x3 = 0 (1)4x4y3 = 0 (2)
}
y = x3 (1)
x = y3 (2)}
-
ExampleLocate all relative extrema and saddle points of f (x, y)= 4xy x4 y4.
fx(x, y)= 4y 4x3 and fy (x, y)= 4x4y3
Critical Points:
4y 4x3 = 0 (1)4x4y3 = 0 (2)
} y = x
3 (1)x = y3 (2)
}
-
y = x3 (1)x = y3 (2)
}
Substituting (1) into (2):
x = (x3)3x = x9
x9x = 0x(x81) = 0
x(x41)(x4+1) = 0x(x21)(x2+1)(x4+1) = 0
x(x1)(x+1)(x2+1)(x4+1) = 0
x = 0 x =1 x = 1y = 0 y =1 y = 1
-
y = x3 (1)x = y3 (2)
}Substituting (1) into (2):
x = (x3)3x = x9
x9x = 0x(x81) = 0
x(x41)(x4+1) = 0x(x21)(x2+1)(x4+1) = 0
x(x1)(x+1)(x2+1)(x4+1) = 0
x = 0 x =1 x = 1y = 0 y =1 y = 1
-
y = x3 (1)x = y3 (2)
}Substituting (1) into (2):
x = (x3)3
x = x9x9x = 0
x(x81) = 0x(x41)(x4+1) = 0
x(x21)(x2+1)(x4+1) = 0x(x1)(x+1)(x2+1)(x4+1) = 0
x = 0 x =1 x = 1y = 0 y =1 y = 1
-
y = x3 (1)x = y3 (2)
}Substituting (1) into (2):
x = (x3)3x = x9
x9x = 0x(x81) = 0
x(x41)(x4+1) = 0x(x21)(x2+1)(x4+1) = 0
x(x1)(x+1)(x2+1)(x4+1) = 0
x = 0 x =1 x = 1y = 0 y =1 y = 1
-
y = x3 (1)x = y3 (2)
}Substituting (1) into (2):
x = (x3)3x = x9
x9x = 0
x(x81) = 0x(x41)(x4+1) = 0
x(x21)(x2+1)(x4+1) = 0x(x1)(x+1)(x2+1)(x4+1) = 0
x = 0 x =1 x = 1y = 0 y =1 y = 1
-
y = x3 (1)x = y3 (2)
}Substituting (1) into (2):
x = (x3)3x = x9
x9x = 0x(x81) = 0
x(x41)(x4+1) = 0x(x21)(x2+1)(x4+1) = 0
x(x1)(x+1)(x2+1)(x4+1) = 0
x = 0 x =1 x = 1y = 0 y =1 y = 1
-
y = x3 (1)x = y3 (2)
}Substituting (1) into (2):
x = (x3)3x = x9
x9x = 0x(x81) = 0
x(x41)(x4+1) = 0
x(x21)(x2+1)(x4+1) = 0x(x1)(x+1)(x2+1)(x4+1) = 0
x = 0 x =1 x = 1y = 0 y =1 y = 1
-
y = x3 (1)x = y3 (2)
}Substituting (1) into (2):
x = (x3)3x = x9
x9x = 0x(x81) = 0
x(x41)(x4+1) = 0x(x21)(x2+1)(x4+1) = 0
x(x1)(x+1)(x2+1)(x4+1) = 0
x = 0 x =1 x = 1y = 0 y =1 y = 1
-
y = x3 (1)x = y3 (2)
}Substituting (1) into (2):
x = (x3)3x = x9
x9x = 0x(x81) = 0
x(x41)(x4+1) = 0x(x21)(x2+1)(x4+1) = 0
x(x1)(x+1)(x2+1)(x4+1) = 0
x = 0 x =1 x = 1y = 0 y =1 y = 1
-
y = x3 (1)x = y3 (2)
}Substituting (1) into (2):
x = (x3)3x = x9
x9x = 0x(x81) = 0
x(x41)(x4+1) = 0x(x21)(x2+1)(x4+1) = 0
x(x1)(x+1)(x2+1)(x4+1) = 0
x = 0 x =1 x = 1y = 0 y =1 y = 1
-
y = x3 (1)x = y3 (2)
}Substituting (1) into (2):
x = (x3)3x = x9
x9x = 0x(x81) = 0
x(x41)(x4+1) = 0x(x21)(x2+1)(x4+1) = 0
x(x1)(x+1)(x2+1)(x4+1) = 0
x = 0
x =1 x = 1y = 0 y =1 y = 1
-
y = x3 (1)x = y3 (2)
}Substituting (1) into (2):
x = (x3)3x = x9
x9x = 0x(x81) = 0
x(x41)(x4+1) = 0x(x21)(x2+1)(x4+1) = 0
x(x1)(x+1)(x2+1)(x4+1) = 0
x = 0 x =1
x = 1y = 0 y =1 y = 1
-
y = x3 (1)x = y3 (2)
}Substituting (1) into (2):
x = (x3)3x = x9
x9x = 0x(x81) = 0
x(x41)(x4+1) = 0x(x21)(x2+1)(x4+1) = 0
x(x1)(x+1)(x2+1)(x4+1) = 0
x = 0 x =1 x = 1
y = 0 y =1 y = 1
-
y = x3 (1)x = y3 (2)
}Substituting (1) into (2):
x = (x3)3x = x9
x9x = 0x(x81) = 0
x(x41)(x4+1) = 0x(x21)(x2+1)(x4+1) = 0
x(x1)(x+1)(x2+1)(x4+1) = 0
x = 0 x =1 x = 1y = 0
y =1 y = 1
-
y = x3 (1)x = y3 (2)
}Substituting (1) into (2):
x = (x3)3x = x9
x9x = 0x(x81) = 0
x(x41)(x4+1) = 0x(x21)(x2+1)(x4+1) = 0
x(x1)(x+1)(x2+1)(x4+1) = 0
x = 0 x =1 x = 1y = 0 y =1
y = 1
-
y = x3 (1)x = y3 (2)
}Substituting (1) into (2):
x = (x3)3x = x9
x9x = 0x(x81) = 0
x(x41)(x4+1) = 0x(x21)(x2+1)(x4+1) = 0
x(x1)(x+1)(x2+1)(x4+1) = 0
x = 0 x =1 x = 1y = 0 y =1 y = 1
-
First Partial Derivatives:
fx(x, y)= 4y 4x3 and fy (x, y)= 4x4y3
Critical Points: (0,0), (1,1), (1,1)C.P. (0,0) (1,1) (1,1)
fxx(x, y)=12x2
0 12 12
fy y (x, y)=12y2
0 12 12
fxy (x, y)= 4
4 4 4
D
16 128 128
conclusion
saddle pt rel max rel max
-
First Partial Derivatives:
fx(x, y)= 4y 4x3 and fy (x, y)= 4x4y3
Critical Points: (0,0), (1,1), (1,1)C.P. (0,0) (1,1) (1,1)
fxx(x, y)=12x2 0
12 12
fy y (x, y)=12y2
0 12 12
fxy (x, y)= 4
4 4 4
D
16 128 128
conclusion
saddle pt rel max rel max
-
First Partial Derivatives:
fx(x, y)= 4y 4x3 and fy (x, y)= 4x4y3
Critical Points: (0,0), (1,1), (1,1)C.P. (0,0) (1,1) (1,1)
fxx(x, y)=12x2 0 12
12
fy y (x, y)=12y2
0 12 12
fxy (x, y)= 4
4 4 4
D
16 128 128
conclusion
saddle pt rel max rel max
-
First Partial Derivatives:
fx(x, y)= 4y 4x3 and fy (x, y)= 4x4y3
Critical Points: (0,0), (1,1), (1,1)C.P. (0,0) (1,1) (1,1)
fxx(x, y)=12x2 0 12 12fy y (x, y)=12y2
0 12 12
fxy (x, y)= 4
4 4 4
D
16 128 128
conclusion
saddle pt rel max rel max
-
First Partial Derivatives:
fx(x, y)= 4y 4x3 and fy (x, y)= 4x4y3
Critical Points: (0,0), (1,1), (1,1)C.P. (0,0) (1,1) (1,1)
fxx(x, y)=12x2 0 12 12fy y (x, y)=12y2 0
12 12
fxy (x, y)= 4
4 4 4
D
16 128 128
conclusion
saddle pt rel max rel max
-
First Partial Derivatives:
fx(x, y)= 4y 4x3 and fy (x, y)= 4x4y3
Critical Points: (0,0), (1,1), (1,1)C.P. (0,0) (1,1) (1,1)
fxx(x, y)=12x2 0 12 12fy y (x, y)=12y2 0 12
12
fxy (x, y)= 4
4 4 4
D
16 128 128
conclusion
saddle pt rel max rel max
-
First Partial Derivatives:
fx(x, y)= 4y 4x3 and fy (x, y)= 4x4y3
Critical Points: (0,0), (1,1), (1,1)C.P. (0,0) (1,1) (1,1)
fxx(x, y)=12x2 0 12 12fy y (x, y)=12y2 0 12 12
fxy (x, y)= 4
4 4 4
D
16 128 128
conclusion
saddle pt rel max rel max
-
First Partial Derivatives:
fx(x, y)= 4y 4x3 and fy (x, y)= 4x4y3
Critical Points: (0,0), (1,1), (1,1)C.P. (0,0) (1,1) (1,1)
fxx(x, y)=12x2 0 12 12fy y (x, y)=12y2 0 12 12
fxy (x, y)= 4 4
4 4
D
16 128 128
conclusion
saddle pt rel max rel max
-
First Partial Derivatives:
fx(x, y)= 4y 4x3 and fy (x, y)= 4x4y3
Critical Points: (0,0), (1,1), (1,1)C.P. (0,0) (1,1) (1,1)
fxx(x, y)=12x2 0 12 12fy y (x, y)=12y2 0 12 12
fxy (x, y)= 4 4 4
4
D
16 128 128
conclusion
saddle pt rel max rel max
-
First Partial Derivatives:
fx(x, y)= 4y 4x3 and fy (x, y)= 4x4y3
Critical Points: (0,0), (1,1), (1,1)C.P. (0,0) (1,1) (1,1)
fxx(x, y)=12x2 0 12 12fy y (x, y)=12y2 0 12 12
fxy (x, y)= 4 4 4 4D
16 128 128
conclusion
saddle pt rel max rel max
-
First Partial Derivatives:
fx(x, y)= 4y 4x3 and fy (x, y)= 4x4y3
Critical Points: (0,0), (1,1), (1,1)C.P. (0,0) (1,1) (1,1)
fxx(x, y)=12x2 0 12 12fy y (x, y)=12y2 0 12 12
fxy (x, y)= 4 4 4 4D 16
128 128
conclusion
saddle pt rel max rel max
-
First Partial Derivatives:
fx(x, y)= 4y 4x3 and fy (x, y)= 4x4y3
Critical Points: (0,0), (1,1), (1,1)C.P. (0,0) (1,1) (1,1)
fxx(x, y)=12x2 0 12 12fy y (x, y)=12y2 0 12 12
fxy (x, y)= 4 4 4 4D 16 128
128
conclusion
saddle pt rel max rel max
-
First Partial Derivatives:
fx(x, y)= 4y 4x3 and fy (x, y)= 4x4y3
Critical Points: (0,0), (1,1), (1,1)C.P. (0,0) (1,1) (1,1)
fxx(x, y)=12x2 0 12 12fy y (x, y)=12y2 0 12 12
fxy (x, y)= 4 4 4 4D 16 128 128
conclusion
saddle pt rel max rel max
-
First Partial Derivatives:
fx(x, y)= 4y 4x3 and fy (x, y)= 4x4y3
Critical Points: (0,0), (1,1), (1,1)C.P. (0,0) (1,1) (1,1)
fxx(x, y)=12x2 0 12 12fy y (x, y)=12y2 0 12 12
fxy (x, y)= 4 4 4 4D 16 128 128
conclusion saddle pt
rel max rel max
-
First Partial Derivatives:
fx(x, y)= 4y 4x3 and fy (x, y)= 4x4y3
Critical Points: (0,0), (1,1), (1,1)C.P. (0,0) (1,1) (1,1)
fxx(x, y)=12x2 0 12 12fy y (x, y)=12y2 0 12 12
fxy (x, y)= 4 4 4 4D 16 128 128
conclusion saddle pt rel max
rel max
-
First Partial Derivatives:
fx(x, y)= 4y 4x3 and fy (x, y)= 4x4y3
Critical Points: (0,0), (1,1), (1,1)C.P. (0,0) (1,1) (1,1)
fxx(x, y)=12x2 0 12 12fy y (x, y)=12y2 0 12 12
fxy (x, y)= 4 4 4 4D 16 128 128
conclusion saddle pt rel max rel max
-
Outline
1 Relative Extreme Function Values
2 Absolute Extrema of Functions of Two Variables
-
DefinitionLet f be a function of two variables x and y . Let R be a region on thexy-plane and (x0, y0) a point in R.
1 f is said to have an absolute maximum on R at (x0, y0) iff (x0, y0) f (x, y) for every (x, y) R.
2 f is said to have an absolute minimum on R at (x0, y0) iff (x0, y0) f (x, y) for every (x, y) R.
3 f is said to have an absolute extremum on R at (x0, y0) if either fhas an absolute maximum value or absolute minimum value on R at(x0, y0).
-
DefinitionLet f be a function of two variables x and y . Let R be a region on thexy-plane and (x0, y0) a point in R.
1 f is said to have an absolute maximum on R at (x0, y0) iff (x0, y0) f (x, y) for every (x, y) R.
2 f is said to have an absolute minimum on R at (x0, y0) iff (x0, y0) f (x, y) for every (x, y) R.
3 f is said to have an absolute extremum on R at (x0, y0) if either fhas an absolute maximum value or absolute minimum value on R at(x0, y0).
-
DefinitionLet f be a function of two variables x and y . Let R be a region on thexy-plane and (x0, y0) a point in R.
1 f is said to have an absolute maximum on R at (x0, y0) iff (x0, y0) f (x, y) for every (x, y) R.
2 f is said to have an absolute minimum on R at (x0, y0) iff (x0, y0) f (x, y) for every (x, y) R.
3 f is said to have an absolute extremum on R at (x0, y0) if either fhas an absolute maximum value or absolute minimum value on R at(x0, y0).
-
DefinitionLet f be a function of two variables x and y . Let R be a region on thexy-plane and (x0, y0) a point in R.
1 f is said to have an absolute maximum on R at (x0, y0) iff (x0, y0) f (x, y) for every (x, y) R.
2 f is said to have an absolute minimum on R at (x0, y0) iff (x0, y0) f (x, y) for every (x, y) R.
3 f is said to have an absolute extremum on R at (x0, y0) if either fhas an absolute maximum value or absolute minimum value on R at(x0, y0).
-
DefinitionLet f be a function of two variables x and y . Let R be a region on thexy-plane and (x0, y0) a point in R.
1 f is said to have an absolute maximum on R at (x0, y0) iff (x0, y0) f (x, y) for every (x, y) R.
2 f is said to have an absolute minimum on R at (x0, y0) iff (x0, y0) f (x, y) for every (x, y) R.
3 f is said to have an absolute extremum on R at (x0, y0) if either fhas an absolute maximum value or absolute minimum value on R at(x0, y0).
-
RemarkSuppose that f has an absolute maximum [minimum, extremum] at(x0, y0).
1 We say f (x0, y0) is the absolute maximum [minimum, extremum] valueof f on R.
2 What is the difference between a relative extremum point and anabsolute extremum point?
-
RemarkSuppose that f has an absolute maximum [minimum, extremum] at(x0, y0).
1 We say f (x0, y0) is the absolute maximum [minimum, extremum] valueof f on R.
2 What is the difference between a relative extremum point and anabsolute extremum point?
-
RemarkSuppose that f has an absolute maximum [minimum, extremum] at(x0, y0).
1 We say f (x0, y0) is the absolute maximum [minimum, extremum] valueof f on R.
2 What is the difference between a relative extremum point and anabsolute extremum point?
-
RemarkSuppose that f has an absolute maximum [minimum, extremum] at(x0, y0).
1 We say f (x0, y0) is the absolute maximum [minimum, extremum] valueof f on R.
2 What is the difference between a relative extremum point and anabsolute extremum point?
-
DefinitionLet R be a region in the xy-plane
1 R is bounded if it can be contained in some rectangle.2 R closed if R contains its boundary.
-
DefinitionLet R be a region in the xy-plane
1 R is bounded if it can be contained in some rectangle.
2 R closed if R contains its boundary.
-
DefinitionLet R be a region in the xy-plane
1 R is bounded if it can be contained in some rectangle.2 R closed if R contains its boundary.
-
Theorem (Extreme Value Theorem)If f (x, y) is continuous on a closed and bounded set R, then
f has both anabsolute maximum and an absolute minimum on R.
-
Theorem (Extreme Value Theorem)If f (x, y) is continuous on a closed and bounded set R, then f has both anabsolute maximum and an absolute minimum on R.
-
RemarkIf f and R satisfy the conditions of EVT, the absolute extrema occur at
1 critical points inside R
2 points on the boundary of R
-
RemarkIf f and R satisfy the conditions of EVT, the absolute extrema occur at
1 critical points inside R2 points on the boundary of R
-
ProcedureFinding Absolute Extrema on a Region R satisfying EVT:
1 Find the critical numbers of f (x, y) in R.
Evaluate f (x, y) at thecritical numbers.
2 Evaluate f (x, y) at the vertices of R (if any).3 Convert f (x, y) to a function of one variable g (x) or g
(y)on the
boundaries of R.4 Find the critical numbers of g and evaluate f (x, y) at these numbers.
Recall: For a function of one variable, the critical numbers are pointsfor which the derivative of g is 0 or does not exist.
5 Compare the values obtained.
-
ProcedureFinding Absolute Extrema on a Region R satisfying EVT:
1 Find the critical numbers of f (x, y) in R. Evaluate f (x, y) at thecritical numbers.
2 Evaluate f (x, y) at the vertices of R (if any).3 Convert f (x, y) to a function of one variable g (x) or g
(y)on the
boundaries of R.4 Find the critical numbers of g and evaluate f (x, y) at these numbers.
Recall: For a function of one variable, the critical numbers are pointsfor which the derivative of g is 0 or does not exist.
5 Compare the values obtained.
-
ProcedureFinding Absolute Extrema on a Region R satisfying EVT:
1 Find the critical numbers of f (x, y) in R. Evaluate f (x, y) at thecritical numbers.
2 Evaluate f (x, y) at the vertices of R (if any).
3 Convert f (x, y) to a function of one variable g (x) or g(y)on the
boundaries of R.4 Find the critical numbers of g and evaluate f (x, y) at these numbers.
Recall: For a function of one variable, the critical numbers are pointsfor which the derivative of g is 0 or does not exist.
5 Compare the values obtained.
-
ProcedureFinding Absolute Extrema on a Region R satisfying EVT:
1 Find the critical numbers of f (x, y) in R. Evaluate f (x, y) at thecritical numbers.
2 Evaluate f (x, y) at the vertices of R (if any).3 Convert f (x, y) to a function of one variable g (x) or g
(y)on the
boundaries of R.
4 Find the critical numbers of g and evaluate f (x, y) at these numbers.Recall: For a function of one variable, the critical numbers are pointsfor which the derivative of g is 0 or does not exist.
5 Compare the values obtained.
-
ProcedureFinding Absolute Extrema on a Region R satisfying EVT:
1 Find the critical numbers of f (x, y) in R. Evaluate f (x, y) at thecritical numbers.
2 Evaluate f (x, y) at the vertices of R (if any).3 Convert f (x, y) to a function of one variable g (x) or g
(y)on the
boundaries of R.4 Find the critical numbers of g and evaluate f (x, y) at these numbers.
Recall: For a function of one variable, the critical numbers are pointsfor which the derivative of g is 0 or does not exist.
5 Compare the values obtained.
-
ProcedureFinding Absolute Extrema on a Region R satisfying EVT:
1 Find the critical numbers of f (x, y) in R. Evaluate f (x, y) at thecritical numbers.
2 Evaluate f (x, y) at the vertices of R (if any).3 Convert f (x, y) to a function of one variable g (x) or g
(y)on the
boundaries of R.4 Find the critical numbers of g and evaluate f (x, y) at these numbers.
Recall: For a function of one variable, the critical numbers are pointsfor which the derivative of g is 0 or does not exist.
5 Compare the values obtained.
-
ProcedureFinding Absolute Extrema on a Region R satisfying EVT:
1 Find the critical numbers of f (x, y) in R. Evaluate f (x, y) at thecritical numbers.
2 Evaluate f (x, y) at the vertices of R (if any).3 Convert f (x, y) to a function of one variable g (x) or g
(y)on the
boundaries of R.4 Find the critical numbers of g and evaluate f (x, y) at these numbers.
Recall: For a function of one variable, the critical numbers are pointsfor which the derivative of g is 0 or does not exist.
5 Compare the values obtained.
-
ExampleFind the absolute minimum and maximum of f (x, y)= x2+2y2x in theclosed disk {(x, y) | x2+ y2 4}
fx(x, y)= 2x1 and fy (x, y)= 4yCritical Point:
(12 ,0
)At critical point inside R: f
(12 ,0
)= 14At vertices: noneAt boundary: f
(12 ,
p152
)= f
(12 ,
p152
)= 334
CONCLUSION: f has an absolute minimum at(12 ,0
)and absolute
maxima at(12 ,
p152
)and
(12 ,
p152
)with values 14 and 334 , respectively.
-
ExampleFind the absolute minimum and maximum of f (x, y)= x2+2y2x in theclosed disk {(x, y) | x2+ y2 4}
fx(x, y)=
2x1 and fy (x, y)= 4yCritical Point:
(12 ,0
)At critical point inside R: f
(12 ,0
)= 14At vertices: noneAt boundary: f
(12 ,
p152
)= f
(12 ,
p152
)= 334
CONCLUSION: f has an absolute minimum at(12 ,0
)and absolute
maxima at(12 ,
p152
)and
(12 ,
p152
)with values 14 and 334 , respectively.
-
ExampleFind the absolute minimum and maximum of f (x, y)= x2+2y2x in theclosed disk {(x, y) | x2+ y2 4}
fx(x, y)= 2x1
and fy (x, y)= 4yCritical Point:
(12 ,0
)At critical point inside R: f
(12 ,0
)= 14At vertices: noneAt boundary: f
(12 ,
p152
)= f
(12 ,
p152
)= 334
CONCLUSION: f has an absolute minimum at(12 ,0
)and absolute
maxima at(12 ,
p152
)and
(12 ,
p152
)with values 14 and 334 , respectively.
-
ExampleFind the absolute minimum and maximum of f (x, y)= x2+2y2x in theclosed disk {(x, y) | x2+ y2 4}
fx(x, y)= 2x1 and
fy (x, y)= 4yCritical Point:
(12 ,0
)At critical point inside R: f
(12 ,0
)= 14At vertices: noneAt boundary: f
(12 ,
p152
)= f
(12 ,
p152
)= 334
CONCLUSION: f has an absolute minimum at(12 ,0
)and absolute
maxima at(12 ,
p152
)and
(12 ,
p152
)with values 14 and 334 , respectively.
-
ExampleFind the absolute minimum and maximum of f (x, y)= x2+2y2x in theclosed disk {(x, y) | x2+ y2 4}
fx(x, y)= 2x1 and fy (x, y)=
4y
Critical Point:(12 ,0
)At critical point inside R: f
(12 ,0
)= 14At vertices: noneAt boundary: f
(12 ,
p152
)= f
(12 ,
p152
)= 334
CONCLUSION: f has an absolute minimum at(12 ,0
)and absolute
maxima at(12 ,
p152
)and
(12 ,
p152
)with values 14 and 334 , respectively.
-
ExampleFind the absolute minimum and maximum of f (x, y)= x2+2y2x in theclosed disk {(x, y) | x2+ y2 4}
fx(x, y)= 2x1 and fy (x, y)= 4y
Critical Point:(12 ,0
)At critical point inside R: f
(12 ,0
)= 14At vertices: noneAt boundary: f
(12 ,
p152
)= f
(12 ,
p152
)= 334
CONCLUSION: f has an absolute minimum at(12 ,0
)and absolute
maxima at(12 ,
p152
)and
(12 ,
p152
)with values 14 and 334 , respectively.
-
ExampleFind the absolute minimum and maximum of f (x, y)= x2+2y2x in theclosed disk {(x, y) | x2+ y2 4}
fx(x, y)= 2x1 and fy (x, y)= 4yCritical Point:
(12 ,0
)
At critical point inside R: f(12 ,0
)= 14At vertices: noneAt boundary: f
(12 ,
p152
)= f
(12 ,
p152
)= 334
CONCLUSION: f has an absolute minimum at(12 ,0
)and absolute
maxima at(12 ,
p152
)and
(12 ,
p152
)with values 14 and 334 , respectively.
-
ExampleFind the absolute minimum and maximum of f (x, y)= x2+2y2x in theclosed disk {(x, y) | x2+ y2 4}
fx(x, y)= 2x1 and fy (x, y)= 4yCritical Point:
(12 ,0
)At critical point inside R: f
(12 ,0
)=
14At vertices: noneAt boundary: f
(12 ,
p152
)= f
(12 ,
p152
)= 334
CONCLUSION: f has an absolute minimum at(12 ,0
)and absolute
maxima at(12 ,
p152
)and
(12 ,
p152
)with values 14 and 334 , respectively.
-
ExampleFind the absolute minimum and maximum of f (x, y)= x2+2y2x in theclosed disk {(x, y) | x2+ y2 4}
fx(x, y)= 2x1 and fy (x, y)= 4yCritical Point:
(12 ,0
)At critical point inside R: f
(12 ,0
)= 14
At vertices: noneAt boundary: f
(12 ,
p152
)= f
(12 ,
p152
)= 334
CONCLUSION: f has an absolute minimum at(12 ,0
)and absolute
maxima at(12 ,
p152
)and
(12 ,
p152
)with values 14 and 334 , respectively.
-
ExampleFind the absolute minimum and maximum of f (x, y)= x2+2y2x in theclosed disk {(x, y) | x2+ y2 4}
fx(x, y)= 2x1 and fy (x, y)= 4yCritical Point:
(12 ,0
)At critical point inside R: f
(12 ,0
)= 14At vertices:
noneAt boundary: f
(12 ,
p152
)= f
(12 ,
p152
)= 334
CONCLUSION: f has an absolute minimum at(12 ,0
)and absolute
maxima at(12 ,
p152
)and
(12 ,
p152
)with values 14 and 334 , respectively.
-
ExampleFind the absolute minimum and maximum of f (x, y)= x2+2y2x in theclosed disk {(x, y) | x2+ y2 4}
fx(x, y)= 2x1 and fy (x, y)= 4yCritical Point:
(12 ,0
)At critical point inside R: f
(12 ,0
)= 14At vertices: none
At boundary: f(12 ,
p152
)= f
(12 ,
p152
)= 334
CONCLUSION: f has an absolute minimum at(12 ,0
)and absolute
maxima at(12 ,
p152
)and
(12 ,
p152
)with values 14 and 334 , respectively.
-
ExampleFind the absolute minimum and maximum of f (x, y)= x2+2y2x in theclosed disk {(x, y) | x2+ y2 4}
fx(x, y)= 2x1 and fy (x, y)= 4yCritical Point:
(12 ,0
)At critical point inside R: f
(12 ,0
)= 14At vertices: noneAt boundary: f
(12 ,
p152
)= f
(12 ,
p152
)= 334
CONCLUSION: f has an absolute minimum at(12 ,0
)and absolute
maxima at(12 ,
p152
)and
(12 ,
p152
)with values 14 and 334 , respectively.
-
ExampleFind the absolute minimum and maximum of f (x, y)= x2+2y2x in theclosed disk {(x, y) | x2+ y2 4}
fx(x, y)= 2x1 and fy (x, y)= 4yCritical Point:
(12 ,0
)At critical point inside R: f
(12 ,0
)= 14At vertices: noneAt boundary: f
(12 ,
p152
)= f
(12 ,
p152
)= 334
CONCLUSION: f has an absolute minimum at(12 ,0
)and absolute
maxima at(12 ,
p152
)and
(12 ,
p152
)with values 14 and 334 , respectively.
-
ExampleFind the absolute minimum and maximum of f (x, y)= 3xy 6x3y +7 onthe closed triangular region R with vertices at (0,0), (3,0) and (0,5).
fx(x, y)= 3y 6 and fy (x, y)= 3x3Critical Point: (1,2)At critical point: f (1,2)= 1At vertices:f (0,0)= 7 f (0,5)= 8 f (3,0)= 11
-
ExampleFind the absolute minimum and maximum of f (x, y)= 3xy 6x3y +7 onthe closed triangular region R with vertices at (0,0), (3,0) and (0,5).
fx(x, y)=
3y 6 and fy (x, y)= 3x3Critical Point: (1,2)At critical point: f (1,2)= 1At vertices:f (0,0)= 7 f (0,5)= 8 f (3,0)= 11
-
ExampleFind the absolute minimum and maximum of f (x, y)= 3xy 6x3y +7 onthe closed triangular region R with vertices at (0,0), (3,0) and (0,5).
fx(x, y)= 3y 6
and fy (x, y)= 3x3Critical Point: (1,2)At critical point: f (1,2)= 1At vertices:f (0,0)= 7 f (0,5)= 8 f (3,0)= 11
-
ExampleFind the absolute minimum and maximum of f (x, y)= 3xy 6x3y +7 onthe closed triangular region R with vertices at (0,0), (3,0) and (0,5).
fx(x, y)= 3y 6 and
fy (x, y)= 3x3Critical Point: (1,2)At critical point: f (1,2)= 1At vertices:f (0,0)= 7 f (0,5)= 8 f (3,0)= 11
-
ExampleFind the absolute minimum and maximum of f (x, y)= 3xy 6x3y +7 onthe closed triangular region R with vertices at (0,0), (3,0) and (0,5).
fx(x, y)= 3y 6 and fy (x, y)=
3x3Critical Point: (1,2)At critical point: f (1,2)= 1At vertices:f (0,0)= 7 f (0,5)= 8 f (3,0)= 11
-
ExampleFind the absolute minimum and maximum of f (x, y)= 3xy 6x3y +7 onthe closed triangular region R with vertices at (0,0), (3,0) and (0,5).
fx(x, y)= 3y 6 and fy (x, y)= 3x3
Critical Point: (1,2)At critical point: f (1,2)= 1At vertices:f (0,0)= 7 f (0,5)= 8 f (3,0)= 11
-
ExampleFind the absolute minimum and maximum of f (x, y)= 3xy 6x3y +7 onthe closed triangular region R with vertices at (0,0), (3,0) and (0,5).
fx(x, y)= 3y 6 and fy (x, y)= 3x3Critical Point: (1,2)
At critical point: f (1,2)= 1At vertices:f (0,0)= 7 f (0,5)= 8 f (3,0)= 11
-
ExampleFind the absolute minimum and maximum of f (x, y)= 3xy 6x3y +7 onthe closed triangular region R with vertices at (0,0), (3,0) and (0,5).
fx(x, y)= 3y 6 and fy (x, y)= 3x3Critical Point: (1,2)At critical point: f (1,2)=
1At vertices:f (0,0)= 7 f (0,5)= 8 f (3,0)= 11
-
ExampleFind the absolute minimum and maximum of f (x, y)= 3xy 6x3y +7 onthe closed triangular region R with vertices at (0,0), (3,0) and (0,5).
fx(x, y)= 3y 6 and fy (x, y)= 3x3Critical Point: (1,2)At critical point: f (1,2)= 1
At vertices:f (0,0)= 7 f (0,5)= 8 f (3,0)= 11
-
ExampleFind the absolute minimum and maximum of f (x, y)= 3xy 6x3y +7 onthe closed triangular region R with vertices at (0,0), (3,0) and (0,5).
fx(x, y)= 3y 6 and fy (x, y)= 3x3Critical Point: (1,2)At critical point: f (1,2)= 1At vertices:
f (0,0)= 7 f (0,5)= 8 f (3,0)= 11
-
ExampleFind the absolute minimum and maximum of f (x, y)= 3xy 6x3y +7 onthe closed triangular region R with vertices at (0,0), (3,0) and (0,5).
fx(x, y)= 3y 6 and fy (x, y)= 3x3Critical Point: (1,2)At critical point: f (1,2)= 1At vertices:f (0,0)=
7 f (0,5)= 8 f (3,0)= 11
-
ExampleFind the absolute minimum and maximum of f (x, y)= 3xy 6x3y +7 onthe closed triangular region R with vertices at (0,0), (3,0) and (0,5).
fx(x, y)= 3y 6 and fy (x, y)= 3x3Critical Point: (1,2)At critical point: f (1,2)= 1At vertices:f (0,0)= 7
f (0,5)= 8 f (3,0)= 11
-
ExampleFind the absolute minimum and maximum of f (x, y)= 3xy 6x3y +7 onthe closed triangular region R with vertices at (0,0), (3,0) and (0,5).
fx(x, y)= 3y 6 and fy (x, y)= 3x3Critical Point: (1,2)At critical point: f (1,2)= 1At vertices:f (0,0)= 7 f (0,5)=
8 f (3,0)= 11
-
ExampleFind the absolute minimum and maximum of f (x, y)= 3xy 6x3y +7 onthe closed triangular region R with vertices at (0,0), (3,0) and (0,5).
fx(x, y)= 3y 6 and fy (x, y)= 3x3Critical Point: (1,2)At critical point: f (1,2)= 1At vertices:f (0,0)= 7 f (0,5)= 8
f (3,0)= 11
-
ExampleFind the absolute minimum and maximum of f (x, y)= 3xy 6x3y +7 onthe closed triangular region R with vertices at (0,0), (3,0) and (0,5).
fx(x, y)= 3y 6 and fy (x, y)= 3x3Critical Point: (1,2)At critical point: f (1,2)= 1At vertices:f (0,0)= 7 f (0,5)= 8 f (3,0)=
11
-
ExampleFind the absolute minimum and maximum of f (x, y)= 3xy 6x3y +7 onthe closed triangular region R with vertices at (0,0), (3,0) and (0,5).
fx(x, y)= 3y 6 and fy (x, y)= 3x3Critical Point: (1,2)At critical point: f (1,2)= 1At vertices:f (0,0)= 7 f (0,5)= 8 f (3,0)= 11
-
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary x = 0:
g (y)= f (0, y) = 3y +7g (y) = 3
no critical pointsAt boundary y = 0:
g (x)= f (x,0) = 6x+7g (x) = 6
no critical points
-
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary x = 0:
g (y)=
f (0, y) = 3y +7g (y) = 3
no critical pointsAt boundary y = 0:
g (x)= f (x,0) = 6x+7g (x) = 6
no critical points
-
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary x = 0:
g (y)= f (0, y) =
3y +7g (y) = 3
no critical pointsAt boundary y = 0:
g (x)= f (x,0) = 6x+7g (x) = 6
no critical points
-
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary x = 0:
g (y)= f (0, y) = 3y +7
g (y) = 3
no critical pointsAt boundary y = 0:
g (x)= f (x,0) = 6x+7g (x) = 6
no critical points
-
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary x = 0:
g (y)= f (0, y) = 3y +7g (y) = 3
no critical pointsAt boundary y = 0:
g (x)= f (x,0) = 6x+7g (x) = 6
no critical points
-
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary x = 0:
g (y)= f (0, y) = 3y +7g (y) = 3
no critical points
At boundary y = 0:
g (x)= f (x,0) = 6x+7g (x) = 6
no critical points
-
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary x = 0:
g (y)= f (0, y) = 3y +7g (y) = 3
no critical pointsAt boundary y = 0:
g (x)= f (x,0) = 6x+7g (x) = 6
no critical points
-
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary x = 0:
g (y)= f (0, y) = 3y +7g (y) = 3
no critical pointsAt boundary y = 0:
g (x)=
f (x,0) = 6x+7g (x) = 6
no critical points
-
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary x = 0:
g (y)= f (0, y) = 3y +7g (y) = 3
no critical pointsAt boundary y = 0:
g (x)= f (x,0) =
6x+7g (x) = 6
no critical points
-
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary x = 0:
g (y)= f (0, y) = 3y +7g (y) = 3
no critical pointsAt boundary y = 0:
g (x)= f (x,0) = 6x+7
g (x) = 6
no critical points
-
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary x = 0:
g (y)= f (0, y) = 3y +7g (y) = 3
no critical pointsAt boundary y = 0:
g (x)= f (x,0) = 6x+7g (x) = 6
no critical points
-
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary x = 0:
g (y)= f (0, y) = 3y +7g (y) = 3
no critical pointsAt boundary y = 0:
g (x)= f (x,0) = 6x+7g (x) = 6
no critical points
-
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary y =53x+5:
g (x) = f(x,5
3x+5
)= 3x
(53x+5
)6x3
(53x+5
)+7
= 5x2+24x8g (x) = 10x+24
Critical points:
10x+24= 0 x = 125
& y = 1
f(125 ,1
)= 20.8
-
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary y =53x+5:
g (x) = f(x,5
3x+5
)= 3x
(53x+5
)6x3
(53x+5
)+7
= 5x2+24x8g (x) = 10x+24
Critical points:
10x+24= 0 x = 125
& y = 1
f(125 ,1
)= 20.8
-
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary y =53x+5:
g (x) = f(x,5
3x+5
)
= 3x(53x+5
)6x3
(53x+5
)+7
= 5x2+24x8g (x) = 10x+24
Critical points:
10x+24= 0 x = 125
& y = 1
f(125 ,1
)= 20.8
-
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary y =53x+5:
g (x) = f(x,5
3x+5
)= 3x
(53x+5
)6x3
(53x+5
)+7
= 5x2+24x8g (x) = 10x+24
Critical points:
10x+24= 0 x = 125
& y = 1
f(125 ,1
)= 20.8
-
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary y =53x+5:
g (x) = f(x,5
3x+5
)= 3x
(53x+5
)6x3
(53x+5
)+7
= 5x2+24x8
g (x) = 10x+24
Critical points:
10x+24= 0 x = 125
& y = 1
f(125 ,1
)= 20.8
-
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary y =53x+5:
g (x) = f(x,5
3x+5
)= 3x
(53x+5
)6x3
(53x+5
)+7
= 5x2+24x8g (x) = 10x+24
Critical points:
10x+24= 0 x = 125
& y = 1
f(125 ,1
)= 20.8
-
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary y =53x+5:
g (x) = f(x,5
3x+5
)= 3x
(53x+5
)6x3
(53x+5
)+7
= 5x2+24x8g (x) = 10x+24
Critical points:
10x+24= 0 x = 125
& y = 1
f(125 ,1
)= 20.8
-
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary y =53x+5:
g (x) = f(x,5
3x+5
)= 3x
(53x+5
)6x3
(53x+5
)+7
= 5x2+24x8g (x) = 10x+24
Critical points:
10x+24= 0
x = 125
& y = 1
f(125 ,1
)= 20.8
-
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary y =53x+5:
g (x) = f(x,5
3x+5
)= 3x
(53x+5
)6x3
(53x+5
)+7
= 5x2+24x8g (x) = 10x+24
Critical points:
10x+24= 0 x = 125
&
y = 1
f(125 ,1
)= 20.8
-
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary y =53x+5:
g (x) = f(x,5
3x+5
)= 3x
(53x+5
)6x3
(53x+5
)+7
= 5x2+24x8g (x) = 10x+24
Critical points:
10x+24= 0 x = 125
& y = 1
f(125 ,1
)= 20.8
-
ExampleGiven: f (x, y)= 3xy 6x3y +7
At boundary y =53x+5:
g (x) = f(x,5
3x+5
)= 3x
(53x+5
)6x3
(53x+5
)+7
= 5x2+24x8g (x) = 10x+24
Critical points:
10x+24= 0 x = 125
& y = 1
f(125 ,1
)= 20.8
-
ExampleGiven: f (x, y)= 3xy 6x3y +7
At critical point: f (1,2)= 1At vertices: f (0,0)= 7 f (0,5)=8 f (3,0)=11At boundary y =53x+5: f
(125 ,1
)= 20.8
CONCLUSION: The function has an absolute maximum at(125 ,1
)and an
absolute minimum at (3,0) with values 20.8 and 11, respectively.
-
ExampleGiven: f (x, y)= 3xy 6x3y +7
At critical point: f (1,2)= 1At vertices: f (0,0)= 7 f (0,5)=8 f (3,0)=11At boundary y =53x+5: f
(125 ,1
)= 20.8CONCLUSION: The function has an absolute maximum at
(125 ,1
)and an
absolute minimum at (3,0) with values 20.8 and 11, respectively.
-
RemarkLet f be a function of two variables and R = {(x, y) | x > 0 and y > 0}.Suppose f is continuous on R and (x0, y0) R is the only critical point off .
1 If f (x, y) is very large when either x or y are very close to zero or verylarge, then f has an absolute minimum at (x0, y0).
2 If f (x, y) is "very negative" when either x or y are very close to zeroor very large, then f has an absolute maximum at (x0, y0).
-
RemarkLet f be a function of two variables and R = {(x, y) | x > 0 and y > 0}.Suppose f is continuous on R and (x0, y0) R is the only critical point off .
1 If f (x, y) is very large when either x or y are very close to zero or verylarge, then f has an absolute minimum at (x0, y0).
2 If f (x, y) is "very negative" when either x or y are very close to zeroor very large, then f has an absolute maximum at (x0, y0).
-
RemarkLet f be a function of two variables and R = {(x, y) | x > 0 and y > 0}.Suppose f is continuous on R and (x0, y0) R is the only critical point off .
1 If f (x, y) is very large when either x or y are very close to zero or verylarge, then f has an absolute minimum at (x0, y0).
2 If f (x, y) is "very negative" when either x or y are very close to zeroor very large, then f has an absolute maximum at (x0, y0).
-
RemarkLet f be a function of two variables and R = {(x, y) | x > 0 and y > 0}.Suppose f is continuous on R and (x0, y0) R is the only critical point off .
1 If f (x, y) is very large when either x or y are very close to zero or verylarge, then f has an absolute minimum at (x0, y0).
2 If f (x, y) is "very negative" when either x or y are very close to zeroor very large, then f has an absolute maximum at (x0, y0).
-
ExampleA manufacturer wishes to make an open rectangular box with volume500 cm3 using the least possible amount of material. Find the dimensionsof the box.
Minimize: Surface AreaConstraint: V = 500 cm3ORMinimize: A = lw +2wh+2lhConstraint: lwh = 500ORMinimize: A(l ,w)= lw + 1000
l+ 1000
wRegion: {(l ,w) | l > 0, w > 0}
-
ExampleA manufacturer wishes to make an open rectangular box with volume500 cm3 using the least possible amount of material. Find the dimensionsof the box.
Minimize:
Surface AreaConstraint: V = 500 cm3ORMinimize: A = lw +2wh+2lhConstraint: lwh = 500ORMinimize: A(l ,w)= lw + 1000
l+ 1000
wRegion: {(l ,w) | l > 0, w > 0}
-
ExampleA manufacturer wishes to make an open rectangular box with volume500 cm3 using the least possible amount of material. Find the dimensionsof the box.
Minimize: Surface Area
Constraint: V = 500 cm3ORMinimize: A = lw +2wh+2lhConstraint: lwh = 500ORMinimize: A(l ,w)= lw + 1000
l+ 1000
wRegion: {(l ,w) | l > 0, w > 0}
-
ExampleA manufacturer wishes to make an open rectangular box with volume500 cm3 using the least possible amount of material. Find the dimensionsof the box.
Minimize: Surface AreaConstraint:
V = 500 cm3ORMinimize: A = lw +2wh+2lhConstraint: lwh = 500ORMinimize: A(l ,w)= lw + 1000
l+ 1000
wRegion: {(l ,w) | l > 0, w > 0}
-
ExampleA manufacturer wishes to make an open rectangular box with volume500 cm3 using the least possible amount of material. Find the dimensionsof the box.
Minimize: Surface AreaConstraint: V = 500 cm3
ORMinimize: A = lw +2wh+2lhConstraint: lwh = 500ORMinimize: A(l ,w)= lw + 1000
l+ 1000
wRegion: {(l ,w) | l > 0, w > 0}
-
ExampleA manufacturer wishes to make an open rectangular box with volume500 cm3 using the least possible amount of material. Find the dimensionsof the box.
Minimize: Surface AreaConstraint: V = 500 cm3ORMinimize: A = lw +2wh+2lhConstraint: lwh = 500
ORMinimize: A(l ,w)= lw + 1000
l+ 1000
wRegion: {(l ,w) | l > 0, w > 0}
-
ExampleA manufacturer wishes to make an open rectangular box with volume500 cm3 using the least possible amount of material. Find the dimensionsof the box.
Minimize: Surface AreaConstraint: V = 500 cm3ORMinimize: A = lw +2wh+2lhConstraint: lwh = 500ORMinimize: A(l ,w)= lw + 1000
l+ 1000
wRegion: {(l ,w) | l > 0, w > 0}
-
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Al (l ,w)=w 1000
l2and Aw (l ,w)= l 1000
w2
Critical Point: l = 10 and w = 10.
-
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Al (l ,w)=
w 1000l2
and Aw (l ,w)= l 1000w2
Critical Point: l = 10 and w = 10.
-
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Al (l ,w)=w
1000l2
and Aw (l ,w)= l 1000w2
Critical Point: l = 10 and w = 10.
-
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Al (l ,w)=w 1000
l2
and Aw (l ,w)= l 1000w2
Critical Point: l = 10 and w = 10.
-
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Al (l ,w)=w 1000
l2and
Aw (l ,w)= l 1000w2
Critical Point: l = 10 and w = 10.
-
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Al (l ,w)=w 1000
l2and Aw (l ,w)=
l 1000w2
Critical Point: l = 10 and w = 10.
-
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Al (l ,w)=w 1000
l2and Aw (l ,w)= l
1000w2
Critical Point: l = 10 and w = 10.
-
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Al (l ,w)=w 1000
l2and Aw (l ,w)= l 1000
w2
Critical Point: l = 10 and w = 10.
-
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Al (l ,w)=w 1000
l2and Aw (l ,w)= l 1000
w2
Critical Point: l = 10 and w = 10.
-
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Note that A(l ,w) is continuous on the R.Hold w constant:
liml
lw + 1000l
+ 1000w
=
liml0+
lw + 1000l
+ 1000w
=
Hold l constant:limw lw +
1000
l+ 1000
w=
limw0+
lw + 1000l
+ 1000w
=
Thus A(l ,w) has an absolute minimum at l = 10 and w = 10.The minimum surface area is 300cm2.
-
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Note that A(l ,w) is continuous on the R.
Hold w constant:liml
lw + 1000l
+ 1000w
=
liml0+
lw + 1000l
+ 1000w
=
Hold l constant:limw lw +
1000
l+ 1000
w=
limw0+
lw + 1000l
+ 1000w
=
Thus A(l ,w) has an absolute minimum at l = 10 and w = 10.The minimum surface area is 300cm2.
-
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Note that A(l ,w) is continuous on the R.Hold w constant:
liml
lw + 1000l
+ 1000w
=
liml0+
lw + 1000l
+ 1000w
=
Hold l constant:limw lw +
1000
l+ 1000
w=
limw0+
lw + 1000l
+ 1000w
=
Thus A(l ,w) has an absolute minimum at l = 10 and w = 10.The minimum surface area is 300cm2.
-
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Note that A(l ,w) is continuous on the R.Hold w constant:
liml
lw + 1000l
+ 1000w
=
liml0+
lw + 1000l
+ 1000w
=
Hold l constant:limw lw +
1000
l+ 1000
w=
limw0+
lw + 1000l
+ 1000w
=
Thus A(l ,w) has an absolute minimum at l = 10 and w = 10.The minimum surface area is 300cm2.
-
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Note that A(l ,w) is continuous on the R.Hold w constant:
liml
lw + 1000l
+ 1000w
=
liml0+
lw + 1000l
+ 1000w
=
Hold l constant:limw lw +
1000
l+ 1000
w=
limw0+
lw + 1000l
+ 1000w
=
Thus A(l ,w) has an absolute minimum at l = 10 and w = 10.The minimum surface area is 300cm2.
-
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Note that A(l ,w) is continuous on the R.Hold w constant:
liml
lw + 1000l
+ 1000w
=
liml0+
lw + 1000l
+ 1000w
=
Hold l constant:limw lw +
1000
l+ 1000
w=
limw0+
lw + 1000l
+ 1000w
=
Thus A(l ,w) has an absolute minimum at l = 10 and w = 10.The minimum surface area is 300cm2.
-
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Note that A(l ,w) is continuous on the R.Hold w constant:
liml
lw + 1000l
+ 1000w
=
liml0+
lw + 1000l
+ 1000w
=
Hold l constant:limw lw +
1000
l+ 1000
w=
limw0+
lw + 1000l
+ 1000w
=
Thus A(l ,w) has an absolute minimum at l = 10 and w = 10.The minimum surface area is 300cm2.
-
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Note that A(l ,w) is continuous on the R.Hold w constant:
liml
lw + 1000l
+ 1000w
=
liml0+
lw + 1000l
+ 1000w
=
Hold l constant:
limw lw +
1000
l+ 1000
w=
limw0+
lw + 1000l
+ 1000w
=
Thus A(l ,w) has an absolute minimum at l = 10 and w = 10.The minimum surface area is 300cm2.
-
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Note that A(l ,w) is continuous on the R.Hold w constant:
liml
lw + 1000l
+ 1000w
=
liml0+
lw + 1000l
+ 1000w
=
Hold l constant:limw lw +
1000
l+ 1000
w=
limw0+
lw + 1000l
+ 1000w
=
Thus A(l ,w) has an absolute minimum at l = 10 and w = 10.The minimum surface area is 300cm2.
-
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Note that A(l ,w) is continuous on the R.Hold w constant:
liml
lw + 1000l
+ 1000w
=
liml0+
lw + 1000l
+ 1000w
=
Hold l constant:limw lw +
1000
l+ 1000
w=
limw0+
lw + 1000l
+ 1000w
=
Thus A(l ,w) has an absolute minimum at l = 10 and w = 10.The minimum surface area is 300cm2.
-
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Note that A(l ,w) is continuous on the R.Hold w constant:
liml
lw + 1000l
+ 1000w
=
liml0+
lw + 1000l
+ 1000w
=
Hold l constant:limw lw +
1000
l+ 1000
w=
limw0+
lw + 1000l
+ 1000w
=
Thus A(l ,w) has an absolute minimum at l = 10 and w = 10.The minimum surface area is 300cm2.
-
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Note that A(l ,w) is continuous on the R.Hold w constant:
liml
lw + 1000l
+ 1000w
=
liml0+
lw + 1000l
+ 1000w
=
Hold l constant:limw lw +
1000
l+ 1000
w=
limw0+
lw + 1000l
+ 1000w
=
Thus A(l ,w) has an absolute minimum at l = 10 and w = 10.The minimum surface area is 300cm2.
-
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Note that A(l ,w) is continuous on the R.Hold w constant:
liml
lw + 1000l
+ 1000w
=
liml0+
lw + 1000l
+ 1000w
=
Hold l constant:limw lw +
1000
l+ 1000
w=
limw0+
lw + 1000l
+ 1000w
=
Thus A(l ,w) has an absolute minimum at l = 10 and w = 10.
The minimum surface area is 300cm2.
-
Example
Minimize: A(l ,w)= lw + 1000l
+ 1000w
Region: l > 0, w > 0
Note that A(l ,w) is continuous on the R.Hold w constant:
liml
lw + 1000l
+ 1000w
=
liml0+
lw + 1000l
+ 1000w
=
Hold l constant:limw lw +
1000
l+ 1000
w=
limw0+
lw + 1000l
+ 1000w
=
Thus A(l ,w) has an absolute minimum at l = 10 and w = 10.The minimum surface area is 300cm2.
-
AnnouncementsRecall Class Policy:
1 Missed midterm, prefinal or final exam, with valid excuse: SpecialExam (Harder than the usual)
2 Missed Chapter Quiz, with valid excuse: 1st missed chapter quiz willnot be included in total, succeeding missed cha