Optimization (minimization and maximization) - (3.7)
Review the Extreme Value Theorem:
Let f be continuous on a, b . Then f attains both an absolute maximum and an absolute minimum.
U
Note that if f is not a constant function and there is only one c in a, b such that f Ÿc 0 then fŸc is
either an absolute maximum or absolute minimum on a, b .
Applications in Minimization and Maximization:
1. Set up a mathematical model for a given minimization or maximization problem:
a. independent variable, domain of each independent variable
b. relation of independent variable and dependent variable
2. Solve the problem.
Example Suppose the area of a rectangle is A. Find the dimensions of the rectangle with minimum
perimeter. Find the minimum perimeter.
1. Let l and w be the length and width of the rectangle.
A and the domain of w: w 0.
a. Since lw A, l w
2A 2w 2
b. Relation of the perimeter P of the rectangle and w : PŸw 2l 2w 2A
2w
w
w
2. Minimize PŸw 2A
2w.
w
Find P U Ÿw and w where P U Ÿw 0 :
2
P U Ÿw " 2A2 2 2 "A 2 w
0, w 2 " A 0, w o A
w
w
Consider only w A . Compute P UU Ÿw and check sign of P UU A :
P UU Ÿw 6A3 , P UU A 6A 0.
w
A A
So, P A is the absolute minimum of P. When w A , then l A A . Hence, the rectangle
with minimum perimeter is a square with area A.
A
Example A farmer has 2400 ft of fencing and wants to fence off a rectangular field that borders a straight
river. He needs no fence along the river. What are the dimensions of the field that has the largest
area?
1. Let w and h be the width and the height of the field. Then
a. the total fencing material 2w h 2400, h 2400 " 2w 2Ÿ1200 " w , 0 t w t 1200;
b. the area A wh 2wŸ1200 " w , 0 t w t 1200
2. Maximize A.
U
Compute A Ÿw 2Ÿ1200 " w wŸ"1 4Ÿ600 " w U
Find zeros of A U Ÿw : A Ÿw 0, w 600
Compute A UU Ÿw and check the sign of A UU at w 600 :
A UU Ÿw "4, A UU Ÿ600 0
So, AŸ600 720 000 is the maximum area of A. Check the graph of AŸw :
1
00000
00000
00000
00000
00000
00000
00000
0
200
400
600
800
1000
1200
Example A square sheet of cardboard 18 inches on a side is make into an open box (no top) by cutting
squares of equal size out of each corner and then folding up the sides. Find the dimensions of the
box with the maximum volume.
1. Let the size of the squares being cut be x.
a. Then 0 2x 18, 0 x 9.
b. VŸx lwh Ÿ18 " 2x 2 x 4Ÿ9 " x 2 x, 0 x 9.
2. Find x so that VŸx is maximized.
Compute V U Ÿx :
U
V Ÿx 4 2Ÿ9 " x Ÿ"1 x Ÿ9 " x 2
4Ÿ9 " x Ÿ"2x 9 " x 12Ÿ9 " x Ÿ3 " x U
Find zeros of V Ÿx : x 3, x 9.
UU
UU
Compute V Ÿx and check the sign of V Ÿx at x 3 :
UU
V Ÿx 12Ÿ"Ÿ3 " x " Ÿ9 " x 12Ÿ"12 2x UU
V Ÿ3 12Ÿ"12 6 0
So, VŸ3 4Ÿ6 2 Ÿ3 432 inch 3 is the maximum volume. Check with the graph of V :
100
80
60
40
20
0
2
4 x
6
8
y VŸx Example Find the point the parabola y 9 " x 2 closest to the point Ÿ3, 9 .
1. Let Ÿx, y be a point on the parabola y 9 " x 2 . Then x, y
x, 9 " x 2 and the point 3, 9 is D :
DŸx Ÿx " 3 2 Ÿ9 " x 2 " 9 2
x, 9 " x 2 and the distance between
Ÿx " 3 2 x 4 ,
". x .
2. Minimizing DŸx is equivalent to minimizing fŸx D 2 Ÿx Ÿx " 3 2 x 4 , " . x .
2
U
Compute f Ÿx 2Ÿx " 3 4x 3 2Ÿ2x 3 x " 3 U
Solve zeros of f Ÿx 0, 2x 3 x " 3 0, Solution is: £x 1. 0¤
UU
UU
Compute f Ÿx and check the sign of f Ÿx at x 1 :
UU
UU
f Ÿx 2Ÿ6x 2 1 , f Ÿ1 0.
So, DŸ1 Ÿ1 " 3 2 1 4 5 is the minimum distance from the parabola to the point 3, 9 . Check
with the graph of DŸx and the parabola:
8
8
6
7
4
2
6
-4
5
0
-2
2t
4
-4
4
-6
3
-1
-2
-8
0
1
x
y DŸx 2
3
y 9 " x2
Example Find the dimensions of a 12 ounces soda can with the minimum amount of material.
1. Consider the soda can as a right circular cylinder. Let r and h be the radius and height of the cylinder.
a. =r 2 h 12 1. 80469 inches 3 21. 65628 inches 3 , h 21. 65628
=r 2
b. total materials surface area of the can
, r0
SŸr 2Ÿ=r 2 2=rh 2=r 2 2=r 122 2 =r 2 21. 65628
r
=r
, r0
2. Minimize SŸr 2 =r 2 21. 65628
r
3
U
65628 .
Compute S Ÿr 2 2=r " 21. 65628
2 2=r " 21.
2
r
r2
Find zeros of S : 2=r 3 " 21. 65628 0, r 3 21. 65628 1. 510 548 463 X 1. 51 inches
2=
UU
UU
Compute S and check the sign of S at r 1. 51
2Ÿ21. 65628 S UU Ÿr 2 2= ,
S UU Ÿ1. 51 0
r3
So, SŸ1. 51 2 =Ÿ1. 51 2 21. 65628 43. 010 105 39 inches 2 is the minimum surface area of the can.
1. 51
When r 1. 51, h 21. 656282 3. 023 291 971 X 3. 02 inch. Check with the graph of S :
=Ÿ1. 51 3
80
60
40
20
0
1
2
r
3
y SŸr 4
4
5