Optimization Problems
1. Express the number 10 as a sum of two
nonnegative numbers whose product is as large as
possible.
x = 1st number
y = 2nd number
x + y = 10
P=xy
y  10  x
dp
 10  2 x
dx
0  10  2 x
P  x(10  x)
P  10 x  x 2
+
0
-
5
The number 5 and 5 will make the product a maximum because
the slope goes from positive to negative when x = 5
Optimization Problems
2. A rectangular field is bounded by a fence on 3
sides and a straight stream on the fourth side. Find
the dimensions of the field with the maximum area
that can be enclosed with 1000 feet of fence.
x = width
y = length
x  250 ft
y  1000  2 x
A  x(1000  2 x)
2x + y = 1000 A  1000 x  2 x
dp
A=xy
 1000  4 x
2
x
x
d2p
 4  0
2
dx
100 - 2x
dx
0  1000  4 x
Since the second derivative is negative at a critical point,
250 ft x 500 ft will be the maximum area.
Optimization Problems
3. Find the dimensions of the biggest rectangle that
can be inscribed in the right triangle with
dimensions of 6 cm, 8 cm, and 10 cm.
x = width
y = length
A=xy
3 

A  6  yy
4 

3 2
A  6y  y
4
dA
3
 6 y
dy
2
3
0  6 y
2
y  4 cm
x  3 cm
y4
8-y
8 cm
y
x
y
x
2
d A
3
 0
2
dy
2
Since the second derivative is negative at a
critical point, 6 cm2 will be the maximum area.
10 cm
6 cm
8 y 8 4
 
x
6 3
4 x  24  3 y
3
x  6 y
4
Optimization Problems
4. A piece of wire of 12 cm can be bent into a circle, a
square, or cut into two pieces to make both a square
and a circle. How should the wire be cut to maximize
and minimize the area?
x
C  2 r
x  2 r
x
r
2
12  x
4
12  x
4
A   r 2  s2
 x   12  x 
A 
 

 2   4 
2
12 - x
2
x 2 144  24 x  x 2
A

4
16
dA x
 24  2 x


dx 2
16
Optimization Problems
4. A piece of wire of 12 cm can be bent into a circle, a
square, or cut into two pieces to make both a square
and a circle. How should the wire be cut to
maximize and minimize the area?
x
12 - x
x
 x  12

2
8
8 x  24  2 x
dA x
 24  2 x


dx 2
16
x
 24  2 x
0

2
16
8 x  2 x  24
x(8  2 )  24
12
x
 4
x
0
A
9 cm2
12
cm 5.04 cm2
 4
12 cm
11.46 cm2
Optimization Problems
5. Find the length of the sides of an isosceles triangle
with perimeter of 12 cm whose area is a maximum.
h 2  (6  x ) 2  x 2
x
x
h
x
h 2  x 2  (36  12 x  x 2 )
h
12 – 2x
h 2  x 2  (6  x ) 2
6–x
1
A  bh
2
1
A  (12  2 x)(12 x  36)1/ 2
2
h 2  12 x  36
h  (12 x  36)1/ 2
A  (6  x)(12 x  36)1/ 2
Optimization Problems
5. Find the length of the sides of an isosceles triangle
with perimeter of 12 cm whose area is a maximum.
x
x
h
x
dA  18 x  72

dx
12 x  36
h
12 – 2x
dA
 (12 x  36) 1/ 2 (36  6 x)  (12 x  36) 
dx
6–x
x4
A  (6  x)(12 x  36)1/ 2
dA
1
 (6  x) (12 x  36) 1/ 2 (12)  (12 x  36)1/ 2 (1)
dx
2
+
0
0
 18 x  72
12 x  36
-
4
x = 4 is a maximum by
the 1st derivative test.
The sides are 4 cm x 4
cm x 4 cm
Optimization Problems
6. A triangle is inscribed in a semicircle of radius 10 cm
so that one side is along the diameter. Find the
dimension of the triangle with the maximum area.
b
h
20 cm
1
A  bh
2
b 2  h 2  400
b  400  h 2
1
A  (400  h 2 )1/ 2 h
2
dA 1 
1

2 1/ 2
2 1 / 2
 (400  h ) (1)  h  (400  h ) (2h)
dh 2 
2


dA 1
 (400  h 2 ) 1/ 2 (400  h 2  h 2 )
dh 2

Optimization Problems
6. A triangle is inscribed in a semicircle of radius 10 cm
so that one side is along the diameter. Find the
dimension of the triangle with the maximum area.
b

dA 1
 (400  h 2 ) 1/ 2 (400  2h 2 )
dh 2
h
20 cm

dA 1
 (400  h 2 ) 1/ 2 (400  h 2  h 2 )
dh 2

dA 1
 (400  h 2 ) 1/ 2 (400  2h 2 )
dh 2


0  400  2h 2
h  200  10 2
b  10 2
h is a maximum by the 1st
derivative test.

Optimization Problems
7. A cattle rancher allows 20 bulls per acre. On the average, the
bulls weigh 2000 lb. Estimated show that the average market
per bull will decrease by 50 lbs for each additional bull. How
many bulls per acre should be allowed in order maximize the
total market weight?
dW
Number of bulls  20  x
 1000  100 x
dx
Weight per bull  (2000  50 x)
0  1000 100x
W  (2000  50 x)( 20  x)
W  40000  1000 x  2000 x  50 x 2
W  40000  1000 x  50 x 2
x  10
d 2W
 100  0
2
dx
x is a maximum by the 2nd
derivative test so the farmer
should allow 30 bulls per
acre.
Optimization Problems
8. How should two non-negative numbers be chosen
that their sum is 1 and the sum of their squares is (a)
as large as possible, and (b) as small as possible?
First number  x
Second number  1  x
S  x 2  (1  x) 2
S  x2 1 2x  x2
S  2x  2x 1
2
dS
d 2S
 4x  2
40
2
dx
dx
1
minimum
2
0,1 endpoints
x
x
sum
0 1
1/2 1/2
1
1
0, 1 is the maximum sum
½ , ½ is the minimum sum