Solutions.
1.
The rectangle can be expressed as a function of x, where the height is 12  x 2 and the
base is 2x . The area is A  2 x 12  x 2  24 x  2 x 3 . Now take the derivative:


A  24  6 x . Set the derivative equal to zero and solve: 24  6 x 2  0  x  2 . A
2
negative answer doesn’t make sense, so we use the solution x  2 . Verify that 2 gives a
maximum by using the second derivative test: A  12x , A2  122  24  0 .
Because the second derivative is negative (and the first derivative is zero) when x  2 ,
we know that x  2 produces a maximum. To find the maximum area, plug x  2 into
the area formula: A2  242  22  32 .
3
2.
Draw a picture.
12  2 x
9  2x
x
After we cut out the squares of side x and fold up the sides, the dimensions of the box
will be:
Width:
9  2x
9  2x  0
Length:
12  2 x
12  2 x  0
Height:
x
x0


x  4.5
x6
Using the formula for volume of a rectangular prism, we get an equation for the volume
of the box in terms of x:
Volume:
V  x9  2 x 12  2 x   4 x 3  42 x 2  108x .
1
Take the derivative and set it equal to zero:
V   12 x 2  84 x  108
12 x 2  84 x  108  0
To solve for x, divide by 12 and then use the quadratic formula.
12 x 2  84 x  108  0  x 2  7 x  9  0  x 
7  13
 5.303, 1.697
2
We can’t cut two squares of length 5.303 inches from each side of length 9 inches, so we
7  13
 1.697 . We verify
2
that this value produces a maximum by using the second derivative test:
get rid of that answer. Therefore, the answer must be x 
V   24x  84
V 1.698  241.697  84  43.272  0
Because the second derivative is negative (and the first derivative is zero), the value is a
maximum when x 
3.
7  13
 1.697 .
2
Draw a picture:
y
x
If we call the length of the plot y and the width x, the area of the plot is A  xy  384 .
The perimeter is P  3x  2 y . So, if we want to minimize the length of the fence, we
need to minimize the perimeter of the plot. If we solve the area equation for y, we get
384
y
 384 x 1 . Now we can substitute this for y in the perimeter equation:
x
2


P  3x  2 y  3x  2 384 x 1  3x  768x 1 . Take the derivative of the perimeter
equation: P   3  768 x . Set the derivative equal to zero and solve for x:
2
3  768 x 2  0
Divide by 3
1  256 x 2  0
Rewrite the negative exponent
256
0
x2
Subtract 1 from both sides
256
 1
x2
Multiply both sides by  x 2
1

256  x 2
Take square roots
x  16
Ignore the negative value since it makes no sense
x  16
Now solve for y
y
384 384

 24
x
16
Verify that x  16 produces a minimum by taking the second derivative of the perimeter
equation:
P  
1536
x3
P 16 
1536
16
3

3
0
8
Because the second derivative is positive (and the first derivative is zero), the value
x  16 produces a minimum. Therefore, the minimum amount of fencing material
needed will occur when the dimensions of the outer rectangle are 16 meters by 24 meters.
4.
The volume of this cylinder is V  r 2 h  512 . The material for the can is the surface
area of the cylinder. The formula for the surface area is S  2r 2  2rh . If we solve the
512
volume equation for h, we get h  2 . We now substitute this value into the area
r
 512 
 512 
2
1
formula: S  2r 2  2rh  2r 2  2r  2   2r 2  2
  2r  1024r
r

r




Taking the derivative of the area gives:
S   4r  1024r 2
3
Setting the derivative equal to zero and solving for r gives:
4r  1024r 2  0
r  256r 2  0
r 
256
0
r2
r 3  256
r2
0
r 3  256  0
r 3  256
r3 
256
r3
256


 4.335
To verify that this value of r produces a minimum, plug r  3
256

into the second
derivative and see that the answer is positive (you should do this, and not take my word
for it). Therefore, the perimeter is a minimum when r  3
5.
256

.
If the swimmer swims the whole distance to the cottage, she will be traveling the entire
time at her slowest speed. If she swims straight to the shore first, minimizing her
swimming distance, she will be maximizing her running distance. Therefore, there
should be some point, somewhere between the cottage and the point on the shore directly
opposite her, where the swimmer should come on land to switch from swimming to
running to get to the cottage in the shortest time.
Picture:
Swimmer
DS
500
DW
1800 – x
x
Cottage
1800
4
So, we let x be the distance from the point on the shore directly opposite the swimmer to
the point where she comes on land. We have two distances to consider. The first is the
diagonal distance that the swimmer swims. This distance, which we’ll call D S , is
DS  500 2  x 2 (by Pythagorean Theorem). The second distance, which we’ll call DW
is simply DW  1800  x . Remember that distance = rate  time? We’ll use this
formula to find the total time the swimmer needs. Here’s how.
distance = rate  time
D = R T
Now divide both sides by R to get
D/R=T
DS
 TS
RS
500 2  x 2
 TS
4
DW
 TW
RW
T  TS  TW
1800  x
 TW
6
T
500 2  x 2 1800  x

4
6
So we need to minimize the total time. Let’s simplify first:
T

1
1
1
500 2  x 2  1800  x   500 2  x 2
4
6
4

1/ 2
 300 
1
x
6
Take the derivative (remembering the chain rule):
T 
T 
T 

 2 x   16
(Simplify)

1
6
(Simplify again)

1
6
(Set equal to zero and solve for x)
1 1
 500 2  x 2
4 2
2x
8 500 2  x 2
x
4 500 2  x 2
x
4 500  x
2
2
x
4 500  x
2
2
1 / 2

1
0
6
(Move the 1/6 to the other side)

1
6
(Cross Multiply)
6 x  4 500 2  x 2
(Divide both sides by 2)
3x  2 500 2  x 2
(Square both sides)
5

9 x 2  4 500 2  x 2

9 x 2  4  500 2  4 x 2
(Distribute)
(Solve for x)
5 x 2  4  500 2
x2 
x
4  500 2
5
2  500
5
 447.214
The swimmer should land 1800  447.214  1,325.786 meters from the cottage.
We can verify that this is a minimum by taking the second derivative, but this would be
messy. It is simpler to use a calculator to check the sign of the derivative at a point on
either side of the answer (the first derivative test), or to graph the equation for the time.
6.
Draw a picture:
We need to find an expression for the distance from the point on the circle to the point
2, 1 and then minimize the distance. If we call the coordinates of the point on the circle
6
x, y  then we can find the distance to 2, 1 by using the distance formula:
2
2
D 2  x  2    y  1 . Next, let’s let L  D 2 and minimize L.
Because x 2  y 2  1 , we can solve for y: y  1  x 2 . Substitute this value of y into the


L: L  x  2   y  1  x  2  1  x 2  2 . Now expand and simplify.
2
2
2
2


L   x  2  1  x 2  1  x 2  4 x  4  1  x 2  2 1  x 2  1  6  4 x  2 1  x 2
2
2
So the equation simplifies to L  6  4 x  2 1  x 2 .
Next we take the derivative:

1
L  4  2  1  x 2
2
  2 x   4 
1 / 2
2x
1 x2
Set the derivative equal to zero and solve:
2x
4
0
1 x2
2x
1 x

2
4
1
2x  4 1  x 2
x  2 1 x2

x2  4 1 x2

x 2  4  4x 2
5x 2  4
x2 
x
4
5
2
5

x
2
5
(If we look at the picture, it is obvious we can exclude the negative answer)
7
Now find the y-coordinate:
2
y  1 x
2

 2 
5 4
1
1
y  1  
 
 

5 5
5
5
 5
 2 1 
,
 . We could verify
The coordinates on the circle that are closest to 2, 1 are 
 5 5
that this is a minimum by taking the second derivative, but that would be messy. It is
simpler to use the calculator to check the sign of the derivative at a point on either side of
the answer (the first derivative test), or to graph the equation for L.
7.
Draw a picture
Call the width of the window 2r and the height h. Notice that 2r is the diameter of the
semicircle.
h
2r
The area of the rectangular portion of the window is 2rh and the perimeter is 2r  2h .
r 2
2r
The area of the semicircular portion of the window is
and the perimeter is
 r.
2
2
Therefore, the area of the window is A  2rh 
r 2
and the perimeter of the window is
2
2r  2h  r  288 . We can use the equation for the perimeter to eliminate a variable
r
from the equation for the area. Let’s isolate h: h  144  r  . Now we can substitute
2
2
r  r
r 2

2
for h in the equation for the area: A  2r 144  r   
.
 288r  2r 
2 2
2

8
Next, we take the derivative:
A  288  4r  r
288
. We can verify that this is a
4
maximum by taking the second derivative: A  4    7  0 . According to the
288
second derivative test, the area is a minimum at r 
.
4
If we set this equal to zero and solve for r, we get r 
8.
 v0 2
Note that v0 and g are constants, so we can re-write the function as R  
 g
v 2
Take the derivative: R    0
 g
to zero and solve for  :
 2v 0 2

 g


 cos2   0



 2v 2
cos2 2   0

 g



 sin 2  .



 cos2  . Set the derivative equal


(Divide both sides by the constant)
cos2   0
Although this has an infinite number of solutions, we are only interested in the solutions

from 0 to
radians. The easiest way to see where cos2   0 is to graph y  cos2  .
2

We see that the derivative is zero when   . Using the graph, we see that the first
4


derivative changes from positive to negative at   , so   provides a maximum.
4
4
9
9.
The equation is given, so all we have to do is take the derivative and set it equal to zero.
P  x 3  48 x 2  720 x  1000
P   3 x 2  96 x  720
3 x 2  96 x  720  0
x 2  32 x  240  0
x  12x  20  0
x  12, x  20
These are the critical values. However, the problem also gave endpoints! Remember, a
maximum or minimum occurs either at an endpoint or at a critical point. We use the
candidate’s test to figure out the maximum.
x
0
12
20
40
P  x 3  48 x 2  720 x  1000
-1000
2456
2200
15000
The maximum profit is $15,000,000,000 ($15 billion).
10.
Find the first derivative and set it equal to zero:
y   3x 2  1

3x 2  1  0

x2 
1
3

x
1
3
We are given the endpoints of the interval, so we use the candidate’s test:
x
-3
1

3
1
3
3
y  x3  x
-24
0.385
-0.385
24
The minimum is -24, which occurs when x  3 . The maximum is 24, which occurs
when x  3 .
10
11.
Draw a picture.
4 in
y
100 in 2
2 in
2 in
x
4 in
Let’s make x be the height of the printed material and y to represent the width of the
printed material. From this picture we can make two equations.
First, the area of the printed material is 100 square inches, so 100 = xy. Another
equation will involve the entire piece of paper including the margins.
Here is another equation: A  x  8 y  4 . We need to solve the first equation for either
x or y and substitute it into the second equation. Let’s solve for y:
100
 100 x 1
x
Now substitute the value of y into the area equation:
100  xy  y 

A  x  8 y  4  x  8 100 x 1  4



A  x  8 100 x 1  4  100  4 x  800 x 1  32
Now FOIL
Simplify
A  132  4 x  800 x 1
Take the derivative and set it equal to zero and solve:
A  4  800 x 2
4  800 x 2  0
x 2 
4
1

800 200
x 2  200

x   200  200  10 2
(Ignore the negative value)
11
To see that x  200 produces a minimum, use the second derivative test.
A  1600 x 3 

1600
x3

1600
A 200 
200
3
The minimum area occurs when x  200 . y 
y
0
100
, so
x
100 100 200
200 10 2



5 2.
200
2
2
200
The dimensions that minimize the material would be 4  5 2 by 8  10 2 , or
approximately 11.1 in by 22.1 in.
12.
Draw a picture:
This time, we’ll let b represent each side of the base and h represent the height.
The volume of the box is V  b 2 h  256 . The area of the box is A  4bh  b 2
256
 256b  2 . Now substitute this expression of h
2
b
2
A  4bh  b  4b 256b 2  b 2  1024b 1  b 2 .
Solve the volume equation for h: h 
into the Area formula:


Take the derivative and set it equal to zero:
A  1024b 2  2b
 1024b 2  2b  0

1024  2b

1
b2
 2b 3  1024
b 3  512
b 8
12
To see if this value is a minimum, use the second derivative test:
A  2048b 3  2 
A8 
Find h: h 
2048
2
b3
2048
260
83
256 256
 2 4
b2
8
The dimensions of the box that require the minimum amount of material is 8 in by 8 in by
4 in.
13.
Draw a picture:
We need to find an expression for the distance from a point on the curve to the point
4, 0 and then minimize the distance. If we call the coordinates of the point on the curve
x, y  then we can find the distance to 4, 0 by using the distance formula:
D 2  x  4   y  0 . Next, let’s let L  D 2 and minimize L.
2
2
13
Because y 
x , we substitute this value of y into the L:
L  x  4 
2
 x .
2
Now expand and simplify.
L  x  4 
2
 x
2
 x 2  8 x  16  x  x 2  7 x  16
So the equation simplifies to L  x 2  7 x  16 .
Next we take the derivative and set equal to zero, and solve:
L  2 x  7  0  x 
7
2
Now find the value of y: y  x 
To verify that x 
7
2
7
produces a minimum, use the second derivative test:
2
L   2
7
L    2  0
2
7
The coordinates of the point at a minimum distance from 4, 0 is  ,
2
7
.
2 
14