3.7 modeling and optimization buffalo bills ranch, north platte, nebraska greg kelly, hanford high...
TRANSCRIPT
3.7 Modeling and Optimization
Buffalo Bill’s Ranch, North Platte, NebraskaGreg Kelly, Hanford High School, Richland, WashingtonPhoto by Vickie Kelly, 1999
A Classic Problem
You have 40 feet of fence to enclose a rectangular garden along the side of a barn. What is the maximum area that you can enclose?
x x
40 2x
40 2A x x
240 2A x x
40 4A x
0 40 4x
4 40x
10x 40 2l x
w x 10 ftw
20 ftl
There must be a local maximum here, since the endpoints are minimums.
A Classic Problem
You have 40 feet of fence to enclose a rectangular garden along the side of a barn. What is the maximum area that you can enclose?
x x
40 2x
40 2A x x
240 2A x x
40 4A x
0 40 4x
4 40x
10x
10 40 2 10A
10 20A
2200 ftA40 2l x
w x 10 ftw
20 ftl
Procedure for Solving Optimization Problems
1. Assign symbols to all given quantities and quantitiesto be determined.
2. Write a primary equation for the quantity to be maximized or minimized.
3. Reduce the primary equation to one having ONEindependent variable. This may involve the use ofsecondary equation.
4. Determine the domain. Make sure it makes sense.5. Determine the max or min by differentiation. (Find the first
derivative and set it equal to zero)6. If the maximum or minimum could be at the endpoints, you
have to check them.
Example: What dimensions for a one liter cylindrical can will use the least amount of material?
We can minimize the material by minimizing the area.
22 2A r rh area ofends
lateralarea
We need another equation that relates r and h:
2V r h
31 L 1000 cm21000 r h
2
1000h
r
22
10 02
02A r r
r
2 20002A r
r
2
20004A r
r
Example 5: What dimensions for a one liter cylindrical can will use the least amount of material?
22 2A r rh area ofends
lateralarea
2V r h
31 L 1000 cm21000 r h
2
1000h
r
22
10 02
02A r r
r
2 20002A r
r
2
20004A r
r
2
20000 4 r
r
2
20004 r
r
32000 4 r
3500r
3500
r
5.42 cmr
2
1000
5.42h
10.83 cmh
A manufacturer wants to design an open box having a square base and a surface area of 108 in2.What dimensions will produce a box with maximumvolume?
x
x
h
Since the box has a squarebase, its volume isV = x2h
Note: We call this the primaryequation because it gives a formula for the quantity we wish to optimize.
The surface area = the area of the base + the area of the 4 sides.
S.A. = x2 + 4xh = 108 We want to maximize the volume,so express it as a function of just one variable. To do this, solvex2 + 4xh = 108 for h.
x
xh
4
108 2 Substitute this into the Volume equation.
x
xxhxV
4
108 222
427
3xx
To maximize V we find the derivative and it’s C.N.’s.
04
327
2
x
dx
dV 3x2 = 108 6.'. xsNC
We can conclude that V is a maximum when x = 6 andthe dimensions of the box are 6 in. x 6 in. x 3 in.
Find the points on the graph of y = 4 – x2 that are closest to the point (0,2).
4
3
2
1
(x,y)(x,y)
What is the distance from the point (x,y) and (0,2)?
22 )2()0( yxd
We want this dist. to be minimized.
43 24 xx
222 )24( xx
21
24
3
432
64'
xx
xxd
21
24
3
432
64'
xx
xxd
imaginary
= 2x(2x2 – 3) = 0
C.N.’s 2
6,0
2
6 02
6
dec.
inc.
dec.
inc.
1st der. test
2
5,
2
6
minimum
2
5,
2
6
minimum
Two closest points.