Transcript OPtimization

Section 5.4
Using Calculus to Solve Optimization Problems
5.3
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1. The sum of two nonnegative numbers is 20. Find the numbers
(a) if the sum of their squares is to be as large as possible.
Let the two numbers be represented by x and 20 – x.
y  x   20  x 
2
2
y '  2x  2  20  x  1
0  4x  40
x  10
f "  x   4 makes x = 10 a minimum.
Maximum must occur at an endpoint.
0 and 20
1. The sum of two nonnegative numbers is 20. Find the numbers
(b) If the product of the square of one number and the
cube of the other is to be as large as possible
Let the two numbers be represented by x and 20 – x.
 20  x 
2
2
y '  3x  20  x   2x 3  20  x  1
y '  x 2  20  x  3  20  x   2x 
0  x 2  20  x  60  5x 
yx
3
2
x  0, 20, 12
_
+
12
+
20
Max at 12, Min at 20
12 and 8
1. The sum of two nonnegative numbers is 20. Find the numbers
(c) if one number plus the square root of the other is as large
as possible.
Let the two numbers be represented by x and 20 – x.
y  20  x  x
y '  1 
0  1 
1
x
2
1
x
4
1
2 x
1
1 3 / 2
y" 
x
4
 1
y "    0 therefore a max
4
2 x
1
and 19 34
4
2. A rectangular pen is to be fenced in using two types of
fencing. Two opposite sides will use heavy duty fencing at $3/ft
while the remaining two sides will use standard fencing at $1/ft.
What are the dimensions of the rectangular plot of greatest area
that can be fenced in at a total cost of $3600?
A  xy
1y
3x
2  3x   2 1y   3600
3x  y  1800
A  x 1800  3x 
A  1800x  3x2
A '  1800  6x
0  1800  6x
300  x
900  y
A "  6
Therefore max
The dimensions of a rectangular plot of greatest area are 300 x 900
3. A rectangular plot is to be bounded on one side by a straight
river and enclosed on the other three sides by a fence. With 800
m of fence at your disposal, what is the largest area you can
enclose?
x
x
y
2x  y  800  y  2x  800
A  xy  A  x  2x  800   A  2x 2  800x
A  2x 2  800x
A '  4x  800
0  4x  800
x  200
A "  4 Therefore a max
A  200  2  200   800   80000
The largest area you can enclose is 80000
4. An open-top box with a square bottom and rectangular sides
is to have a volume of 256 cubic inches. Find the dimensions
that require the minimum amount of material.
S  x  4xy
x
2
 256 
 S  x  4x  2 
y
2
 x 
V  x y  256
x
1024
S  x2 
x
2048
1024
S"  2  3  0
S'  2x  2
x
x
therefore a min
1024
0  2x  2
x
x 8 y4
2
8x8x4
5. Find the largest possible value of 2x + y if x and y are the
lengths of the sides of a right triangle whose hypotenuse is
5 units long.
B  2x  y
x y 5
2
B'  2 
0  2
2
 B  2x  5  x
5
y
2
x
2x
2 5x
x
5  x2
x 2 5x

x2  4
x  2,  2
_
+
2
Therefore x = 2 is a max
22  y 2  5
2
x2  4 5  x2
2

y 1
2x  y  2  2   1  5
6. A right triangle of hypotenuse 5 is rotated about one of its
legs to generate a right circular cone. Find the cone of greatest
volume.
x 2  y 2  52
1
2

V


25

y
y
1
2
3
5
V  x y
y
3
25
1
V
y  y 3
3
3
x
25
25
2
V' 
  y 2
V' 
  y
3
3
25
V "  2y
0
  y 2
3
 5 
 5 
V "
  2 
0
5
3
3




y
3
Therefore max
2
2
2 5
 5 
x 2  25  y 2  x 2  25  

x

5

5


3
3
3
 3


7. Determine the area of the largest rectangle that may be
Inscribed under the curve y  e x on  x, x 
A  2xy
A  2xe x
A '  2e x  2xe x
2
A '  x (1  x)
e
2
0  x (1  x)
e
x 1
+
_
1
Therefore max
2
A  2 1 e 
e
1
8. (calculator required) A poster is to contain 100 square inches of picture surrounded
by a 4 inch margin at the top and bottom and a 2 inch margin on each side. Find the
overall dimensions that will minimize the total area of the poster.
A POS    y  8  x  4 
 100


 8   x  4
 x

400
 132 
 8x
x
2
x
2
4
y
4
A PIC   xy  100
A POS    x  4  y  8 
y  14.142
Since f’ changes
from neg to pos,
we have a
minimum
11.1071 22.142
9. Determine the point on the graph of y  x  2 that is
nearest to the point (5, 0).
D
 x  5   y  0
2
2


D   x  9x  23 
1
D'   x  9x  23   2x  9 
2
D  x  10x  25  x  2
2
1/ 2
2
1/ 2
2
_
+
9/2
Therefore min
1/ 2
2x  9
9
D' 
x
2
2
2 x  9x  23
9 5 
 ,

2
2


10. The graphs of y  25  x 2 , x  0 and y  0 bound a region
in the first quadrant. Find the dimensions of the rectangle of
maximum perimeter that can be inscribed in this region.
P  2x  2y
P  2x  2 25  x 2
1
P'  2  2  25  x 2
2
2x
P'  2 
25  x 2

2x  2 25  x
x2  25  x2
25
2
x 
2
5
x
2
  2x 
1/2
_
+
5
2
Therefore max
2
5
2

5
2
11. (calculator required) Find the dimensions of the rectangle
with maximum area that can be inscribed in a circle of radius 10.
A  4xy
x 2  y 2  100
A  4x 100  x 2
x  7.071  y  7.071
14.142  14.142
Since f’ changes
from pos to neg,
we have a
maximum
12. Suppose that the revenue of a company can be represented with the function
r  x   48x, and the company's cost function is c  x   x3  12x 2  60x, where x
represents thousands of units and revenue and cost are represented in thousands
of dollars. What production level maximizes profit and what is the maximum profit
to the nearest thousand dollars?
P  r x  c x
P   48x    x 3  12x 2  60x 
P  12x  x3  12x2
dP
 3x 2  24x  12
dx
0  3x 2  24x  12
Max at x = 7.46
since f ‘
changes from
pos to neg
x  0.54, 7.46
At 7.46, r = 358.08, c = 194.34, or P = 163.14
Max profit is $163,000 which occurs when 7460 units are made
13. A tank with a rectangular sides is to be open at the top. It is to be
constructed so that its width is 4 m and its volume is 36 cubic meters. If
building the tank costs $10 per square meter for the base and $5 per square
meter for the sides, what is the cost of the least expensive tank?
4xy  36  xy  9
4
C   $10  4y   $5  2  4  x    $5  2  x  y 
9
C  40  40x  10  9 
x
C'  360x 2  40
360
 40  x  3
2
x
720
C"  3  0
x
Therefore min
y
x
y3
C  40
9
 40  3   10  9   $330
3
CALCULATOR REQUIRED
14. Find the minimum distance from the origin to the curve y  ex
 x  0   y  0
2
D

D x e
2
2x
2

1/ 2
x  0.426
Minimum since f ‘ (x) changes from
neg to pos at –0.426
D
 0.426
2
e
2 0.426 

1/ 2
 0.780
15. (calculator required) Consider f  x   12  x 2 for 0  x  2 3. Let
A(t) be the area of the triangle formed by the coordinate axes and the


tangent to the graph of f at the point t, 12  t 2 . For what value of t is
f '  x   2x  f '  t   2t
A(t) a minimum?


y  12  t 2  2t  x  t 


t
If y  0, 0  12  t   2t  x  t   x 
If x  0, y  12  t 2  2t  0  t   y  t 2  12
2
A t 
1
xy
2

t  12
1  t  12  2
 
 t  12 
2  2t 
4t
2


2

2
 12
2t
2
Since A ‘ changes from neg to pos, min area at t = 2
16. Find the maximum distance measured horizontally between
the graphs of f  x   x and g  x   x 2 for 0  x  1.
D y y
D' 
1
2 y
1
1
1
1
 2 y  1  4y  1  y 
4
2 y
_
+
1
4
Therefore max
D
1 1 1
 
4 4 4
17. (calculator required) What is the area of the largest rectangle


that can be inscribed under the graph of y  2cos x for
x ?
2
2
A  2xy  A  4xcos x
Since A ‘ changes from
pos to neg at x = 0.860,
max of A occurs at
x = 0.860
A  4  0.86033358  cos  0.86033358   2.244
18. Consider the set of all right circular cylinders for which the
sum of the height and diameter is 18 inches. What is the radius
of the cylinder with the maximum volume?
h  2r  18
V  r 2h
 h  18  2r  V  r 2 18  2r 
V  18r 2  2r 3
V '  36r  6r 2
0, 6
V '  6r  6  r   r  X
V "  36  12r
V "  6   0  max
19. An isosceles triangle has one vertex at the origin and the other
two at the points where a line parallel to and above the x-axis
intersects the curve f  x   12  x 2. Find the maximum area of the triangle.
A  xy  A  x 12  x 2   A  12x  x 3
A '  12  3x2
A "  6x
12  3x 2
x2
A "  2  0
f  2  12  22  8
A  xy  2  8   16
 max
20. Find the height of the rectangle with largest area that can be
1
1  2x 2
2x
A  2xy  A 
1  2x 2
2 1  2x 2  4x  2x 
A' 
2
2
1  2x
inscribed under the graph of y 

_
1
_
+
2
 max at
A' 
1
2
1
2


2  4x
2
1 2x 
2
2
1



2 1  2x 2
1 2x 
2
1
y

2
2
 1 
1 2 

 2

2
CALCULATOR REQUIRED
21. Find the minimum distance from the origin to the curve
x3  3x 2  4x  12
y
6
D
 x  0
2
 x  3x  4x  12


 0
6


3
2
2
Two possibilities where
D ‘ changes from neg
to pos, denoting a min
D 1.0696716   1.410
D  1.781107   1.899
22. Find the volume of the largest right circular cylinder that can
be inscribed in a sphere of radius 5.
2


h
2
2
V  r h  V    5   h
4

h3
V  25h 
4
3h2
V '  25 
2
4
1 
2
2
R
0.5h
r   h  5
2
3

h
2 
 25
r
4
10
h – height of cylinder
h
r – radius of cylinder
3
R – Given radius of sphere
10
3
500
3h
 10 
3
V

V"  
V "

 0 Therefore a max

3 3
2
2
 3