fx y= − 12x− x 4y 2y ⋅ and locate all absolute...

CALCULUS 4 QUIZ #3 REVIEW Part 2 / SPRING 09

(2.) Determine the following about maxima & minima of functions of 2 variables.

(a.) Complete the square for f x y,( ) 1 2 x⋅− x2− 4 y⋅+ 2 y2⋅−= and locate all absolute maxima & minima..

f x y,( ) 1 x2 2 x⋅+ 1+( ) 2 y2 2 y⋅− 1+( )⋅⎡⎣ ⎤⎦−⋅− 1+ 2+=

f x y,( ) 4 x 1+( )2− 2 y 1−( )2⋅−=

The absolute maximum occurs at 1− 1, 4,( ). There is no absolute minimum.

y

x

z

Absolute Maximum Point

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Since fyy16

112

,⎛⎜⎝

⎞⎟⎠

0< , 16

112

,1

432,⎛⎜

⎝⎞⎟⎠ is a maximum.

Therefore, 16

112

,1

432,⎛⎜

⎝⎞⎟⎠ is a max. or min.

point.

D16

112

,⎛⎜⎝

⎞⎟⎠

1 0>=

Therefore, 0 0, 0,( ) is a saddle point.D 0 0,( ) 1− 0<=

D x y,( ) 6− x⋅( ) 2−( )⋅ 1( )2−= 12 x⋅ 1−=

D x y,( ) fxx x y,( )⋅ fyy⋅ x y,( )⋅ fxy x y,( )⋅⎡⎣ ⎤⎦2−=

fxy 1=fyy 2−=fxx 6− x⋅=

Here are the critical points: 0 0, 0,( ), 16

112

,1

432,⎛⎜

⎝⎞⎟⎠.

y 0=112

,y 3 2 y⋅( )2⋅= 12 y2⋅=

yf∂

∂x 2 y⋅−= 0=

xf∂

∂y 3 x2⋅−= 0=

(b.) Locate all relative maxima, minima, & saddle points for f x y,( ) x y⋅ x3− y2−= .

y

z

x

Saddle & Relative Maximum in close proximity

(c.) Locate all relative maxima, minima, & saddle points for

f x y,( ) x2 y2+2

x y⋅+= .

xfd

d2 x⋅

2

x2 y⋅−= 0=

yfd

d2 y⋅

2

x y2⋅−= 0=

xy

⎛⎜⎝

⎞⎟⎠

3 xy

⎛⎜⎝

⎞⎟⎠

=xy

1−= 1,xy

⎛⎜⎝

⎞⎟⎠

21=

But the signs of "x" and "y" must be the same.

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Accordingly, here are the critical points: 1− 1−, 4,( ), 1 1, 4,( ).

fxx 24

x3 y⋅+= fyy 2

4

x y3⋅+= fxy

2

x2 y2⋅=


D x y,( ) 24

x3 y⋅+⎛

⎜⎝

⎞⎟⎠

24

x y3⋅+⎛

⎜⎝

⎞⎟⎠

⋅2

x2 y2⋅

⎛⎜⎝

⎞⎟⎠

2−=

D 1− 1−,( ) 32 0>= D 1 1,( ) 32 0>=

1 1, 32,( )

Accordingly, both points: 1− 1−, 4,( ), 1 1, 4,( ) must be relative maximums or minimums.

Since fxx 1− 1−,( ) 6 0>= , 1− 1−, 4,( ) is a relative minimum.

Since fyy 1 1,( ) 6 0>= , 1 1, 4,( ) is a relative minimum.

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z

y

x

Relative Minimum Points

(d.) Show that the second partials test provides no information about the critical points of the function f x y,( ) x4 y4−= ; classify all critical points as relative maxima, relative minima, or saddle points.

fx 4 x3⋅= 0= fy 4− y3⋅= 0=

Critical point is 0 0, 0,( ).

fxy 0=fxx 12 x2⋅= fyy 12− y2⋅= of 15


D x y,( ) 12 x2⋅( ) 12− y2⋅( )⋅ 0( )2−= 144− x2⋅ y2⋅=

D 0 0,( ) 0=

Since D 0 0,( ) 0= , the 2nd partials test tells us nothing about the nature of the critical point 0 0, 0,( ).

z x4 y4−=

This surface is a hyperbolic paraboloid. It has a relative maximum at the point along the y-axis and a relative minimum along the x-axis. evidently the point is a saddle point 0 0, 0,( ).

xy

z

Saddle Point @ (0,0,0)

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(e.) Find the absolute extrema of f x y,( ) x y⋅ x− 3 y⋅−= on the closed and bounded triangular region "R" with these vertices: 0 0,( ), 0 4,( ), 5 0,( ).

xf∂

∂y 1−= 0=

yf∂

∂x 3−= 0=

Critical Point: 3 1, 3−,( )

fxx x y,( ) 0= fyy x y,( ) 0= fxy x y,( ) 1=

fxx 3 1,( ) 0= fyy 3 1,( ) 0= fxy 3 1,( ) 1=

In this case, D 0< . This critical point is thus a Saddle Point.

Also, this point lies within the bounded region. Here is why.

0 1 2 3 4 5

1

2

3

4R:Triangular Region

x

y

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This is the equation of the hypotenuse.

y x( ) 445

x⋅−=

y 3( ) 4125

−=85

1> 0>=

Thus, the point lies below the hypoteuse and above the base. Therefore, it is an interior point and thus a candidate for an absolute extremum.

Since " R " is "closed" and bounded, there must exist both an Absolute Minimum & an Absolute Maximimum at Interior Points or on Boundary Points.

Side # 1 : x 0= .

f 0 y,( ) 3− y⋅= No Critical Points.

Side # 2 : y 0= .

f x 0,( ) x−= No Critical Points.

Side # 3 : y 445

x⋅−= .

f x 445

x⋅−⎛⎜⎝

⎞⎟⎠

,⎡⎢⎣

⎤⎥⎦

x 445

x⋅−⎛⎜⎝

⎞⎟⎠

⋅ x− 3 445

x⋅−⎛⎜⎝

⎞⎟⎠

⋅−=

f x 445

x⋅−⎛⎜⎝

⎞⎟⎠

,⎡⎢⎣

⎤⎥⎦

x 3−( ) 445

x⋅−⎛⎜⎝

⎞⎟⎠

⋅ x−=

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xf x 4

45

x⋅−⎛⎜⎝

⎞⎟⎠

,⎡⎢⎣

⎤⎥⎦

dd

445

x⋅−⎛⎜⎝

⎞⎟⎠

45

x 3−( )⋅− 1−=

xf x 4

45

x⋅−⎛⎜⎝

⎞⎟⎠

,⎡⎢⎣

⎤⎥⎦

dd

5.485

x⋅−= 0=

x5 5.4( )⋅

8= 3.375= y 3.375( ) 1.3=

Critical Point : 3.375 1.3, 2.89−,( )

Now we must evaluate f x y,( ) at the three vertices.

f 0 0,( ) 0= f 0 4,( ) 12−= f 5 0,( ) 5−=

Evidently , the two vertices at 0 0, 0,( ) & 0 4, 12−,( ) yield, respectively, an Absolute Maximum & Absolute Minimum for " f x y,( ) " on " R ".

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y

z

x

Key Points in "R"

(f.) Find the absolute extrema of f x y,( ) x2 y2−= on the closed and bounded region "R", where "R" is the diskx2 y2+ 4≤ .

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y x

z

f(x,y) & Region: "R"

f x y,( ) x2 y2−=

xf∂

∂2 x⋅=

yf∂

∂2− y⋅=

x 0= y 0=

There is a horizontal tangent planes at the point: 0 0, 0,( ).

Note that the critical point 0 0, 0,( ) is within the boundary of the circle.

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Now we include the 2 points: 2− 0, 4,( ), 2 0, 4,( ) that we broke the circle into 2 branches.

We get the same value for "x" for the lower branch. Here are the points: 0 2, 4−,( ), 0 2−, 4−,( ).

x 0=x

v x( )dd

4 x⋅=

v x( ) f x yu x( )⋅,⎡⎣ ⎤⎦= x2 4 x2−( )−= 2 x2⋅ 4−=

v x( ) f x yu x( )⋅,⎡⎣ ⎤⎦= x2 4 x2−( )2−=

f x y,( ) x2 y2−=

yl x( ) 4 x2−−=yu x( ) 4 x2−=

We break up the circle into an upper and lower branch and exam each branch separately for extremals.

Therefore, the point: 0 0, 0,( ) is a saddle point.

D 0 0,( ) fxx 0 0,( )⋅ fyy⋅ 0 0,( )⋅ fxy 0 0,( )⋅⎡⎣ ⎤⎦2−= 4− 0<=

fyy 0 0,( )⋅ 2−=fxx 0 0,( )⋅ 2= fxy 0 0,( )⋅ 0=

2yf∂

∂

22−=

2xf∂

∂

22= y x

f∂

∂

∂

∂0=

Here are the candidates: 0 0, 0,( ), 0 2, 4−,( ), 0 2−, 4−,( ), 2− 0, 4,( ), and 2 0, 4,( ).

Comparison of the "z" values of all 5 points reveals that the absolute maximum occurs at the points: 2− 0, 4,( ), 2 0, 4,( ) and the absolute minimum occurs at the points: 0 2−, 4−,( ), 0 2, 4−,( ).

y x

z

Absolute Extrermals in Region "R"

(g.) Show that among all parallelograms with perimeter L, a

square with sides of length L4

has maximum area.

Let the adjacent sides of the parallelogram be "a" and "b" with "θ" as the angle betwee those sides.

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L 2 a⋅ 2 b⋅+=

A a b⋅ sin θ( )⋅=

bL2

a−=

A a θ,( ) aL2

a−⎛⎜⎝

⎞⎟⎠

⋅ sin θ( )⋅=

aA∂

∂

L2

2 a⋅−⎛⎜⎝

⎞⎟⎠

sin θ( )⋅= 0=θ

A∂

∂a

L2

a−⎛⎜⎝

⎞⎟⎠

⋅ cos θ( )⋅= 0=

L2

2 a⋅−⎛⎜⎝

⎞⎟⎠

sin θ( )⋅ 0=L2

a−⎛⎜⎝

⎞⎟⎠

cos θ( )⋅ 0=

The angle θ can not be zero because the resulting parallelogram would degenerate to a line and thus zero area. Thus, the only way to satisfy both equations simultaneously is for their common

factor L2

2 a⋅−⎛⎜⎝

⎞⎟⎠ to vanish.

L2

2 a⋅− 0=

aL4

= bL2

a−=L4

=

Thus, the critical point a a, a2,( ) is a square.

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Aaa 2− sin θ( )⋅=

Aaa a a,( ) 2− sinπ2

⎛⎜⎝

⎞⎟⎠

⋅= 2− 0<=

Therefore, out of all the paralleograms with a fixed perimeter, the square yields the maximum enclosed area.

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fx y= − 12x− x 4y 2y ⋅ and locate all absolute...

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