In this section, we'll use our results on maxima and minima for functions to do word problems which involve finding the largest or smallest value of lengths, areas, volumes, costs, and so on. The hardest part of doing these problems is setting up the appropriate equations; the calculus part is relatively simple.

Some of these problems involve finding an absolute max or min on a closed interval. We know that to do this, we find the critical points on the interval, then test the endpoints and the critical points in the original function.

In some situations, the quantity we're trying to maximize or minimize varies over a non-closed interval. In those cases, the following theorem is often useful.

* Theorem.* Suppose a function f has its second
derivative defined on an interval . Suppose that , and that c is
the only critical point on the interval .

(a) If c is a local max, then it is an absolute max.

(b) If c is a local min, then it is an absolute min.

The key in applying the theorem is that there is only one critical point.

* Example.* A rectangular pen consists of two
identical sections separated by a fence. 120 feet of fence is used
for the outside of the pen and the dividing fence. What dimensions
for each section produce the pen of largest area?

Let each section be x feet in width and y feet in height.

The total area is

Since 120 feet of fence are available,

Plug this into :

The endpoints are given by the extreme cases and ; note that gives

Differentiate:

Set and solve for x:

The maximum occurs when . In this case,

* Example.* A rectangular sheet of cardboard is
15 wide and 24 inches high. Squares with sides of length x are cut
out of each corner. The four resulting tabs are folded up to make a
rectangular box (with no top). Find the value of x which gives the
box of largest volume.

The base of the box is by , and the height is x. Hence, the volume is

The endpoints are (cut no squares out) and (cut halfway from the bottom to the top). Obviously, these endpoints give volume 0, but it means that V is defined on a closed interval. So I just need to test the critical points on the interval and find the one that gives a max.

Differentiate:

The roots are and .

Now is greater than , so it's outside the interval. So I only need to consider the other root .

Thus, maximimizes the volume.

* Example.* A rectangular poster is printed on a
piece of cardboard with area 1200 square inches. The printed region
is a rectangular region centered on the cardboard. There are
unprinted (blank) margins of width 3 inches on the left and right of
the printed region, and width 4 inches on the top and bottom of the
printed region.

What dimensions for the printed region maximize its area?

Suppose the printed area is x by y. Its area is

Since there are margins of width 3 on the left and right, and margins of width 4 on top and bottom, the area of the cardboard is

Solving for y yields

Plug this into the expression for A to obtain

The extreme cases are and . If , then

So the endpoints are and . (These make the area of the printed region 0, so they obviously don't give the maximum!)

Compute the derivative:

Find the critical points:

Since x is a length, it can't be negative, so take the plus sign. This gives

Plug this into to get .

This shows that and maximize the area of the printed region.

* Example.* Find the point on the line which is closest to the point .

The distance from to is

I want the point for which the distance is smallest.

Since smaller numbers give smaller squares and vice versa, I can find
where the * square* of the distance is smallest:

This allows me to remove the square root, and will make differentiation easier.

The line is . Solving for x gives . Plug this into the equation for s:

There are no restrictions on x, so I'll use the Second Derivative Test. Differentiate (noting that the variable is y):

Set to find the critical points:

This gives

Now , so is a local min. Since it's the only critical point, it's an absolute min.

* Example.* The volume of a circular cylinder
(with a top and a bottom) is . What values for
the radius r and the height h give the smallest total surface area
(the area of the side plus the area of the top plus the area of the
bottom)?

The area of the side is , the area of the top is , and the area of the bottom is .

The total surface area is

The volume is

Solving for h gives . Plug this into A:

The only restriction on r is . Hence, r is not restricted to a closedinterval . Therefore, I'll use the Second Derivative Test.

Compute and :

Set and solve for r:

This gives .

Now

Hence, is a local min. Since it's the only critical point, it's an absolute min.

* Example.* A cylindrical can with a top and a
bottom is made with square inches of sheet
metal, with no waste. What dimensions for the radius r and the height
h give the can of largest volume?

The area of the top is , the area of the bottom is , and the area of the side is . So the total area is

Solve the equation for h:

Substitute in V:

is ruled out, because it would cause division by 0 in the equation for h. There is no other restriction on r, except that it should be positive. Since V is not restricted to a closed interval , I'll use the Second Derivative Test.

Compute the derivatives:

Find the critical points:

Since r can't be negative (as it's the radius of a cylinder), I get . Then

The second derivative is

Thus, is a local max. Since it's the only critical point, it's an absolute max.

* Example.* Find the positive number for which
the sum of 5 times the number and 320 times its reciprocal is
smallest.

Let x be the number. The sum of 5 times the number and 320 times its reciprocal is

The only restriction on x is . Since x is not restricted to a closed interval , I'll use the Second Derivative Test.

Differentiate:

Find the critical points by setting :

Since , I get . Plug this into the Second Derivative:

Hence, is a local min. Since it's the only critical point, it's an absolute min.

* Example.* A rectangular box with a square
bottom * and no top* is made with 1200 square
inches of cardboard. What values of the length x of a side of the
bottom and the height y give the box with the largest volume?

The volume is

The area of the 4 sides is , and the area of the bottom is . So

Solving for y gives

Plug this into V and simplify:

Note that , since plugged into gives a contradiction. So the only restriction on x is that .

Since x is not restricted to a closed interval , I'll use the Second Derivative Test.

Compute the derivatives:

Find the critical points:

Since x is a length, it must be positive, so . This gives

Plug into the Second Derivative:

is a local max, but it's the only critical point, so it's an absolute max.

* Example.* A rectangular box with a square
bottom * and no top* has a volume of 2048 cubic
inches. What values of the length x of a side of the bottom and the
height y give the box with the smallest total surface area (the area
of the bottom plus the area of the sides)?

The area of the 4 sides is , and the area of the bottom is . So the total area is

The volume is

Solving for y gives

Plug this into A and simplify:

Note that , since plugged into gives , a contradiction. So the only restriction on x is that .

Since x is not restricted to a closed interval , I'll use the Second Derivative Test.

Compute the derivatives:

Find the critical points:

gives

Plug into the Second Derivative:

is a local min, but it's the only critical point, so it's an absolute min.

* Example.* A rectangular box is made with
cardboard. It has no top, and consists of two identical partitions
separated by a common wall. Each partition has a square bottom which
is x inches by x inches, and the height of the box is y inches. If
the volume of the whole box is 10584 cubic inches, what values for x
and y minimize the total amount of cardboard used (for the bottom,
the four sides, and the divider that separates the partitions)?

The area of the bottom is .

The front and back are each by y, so they have a total area of .

The right side, left side, and the middle partition are each x by y, so they have a total area of .

Therefore, the total area (the total amount of cardboard used) is

The total volume is

Solving this equation for y gives

Plug this into A and simplify:

The extreme case is ruled out, because it causes division by 0 in the equation for A. Thus, x is not restricted to a closed interval , and I'll use the Second Derivative Test.

Differentiate:

Find the critical points by setting and solving for x:

This gives

Do the Second Derivative Test:

Therefore, is a local min. Since it's the only critical point, it's an absolute min.

* Example.* A rectangular box with no top has two
identical partitions separated by a common wall. Each partition has a
square bottom. If 2400 square inches of cardboard are used with no
waste to construct the box, what dimensions yield the box with the
largest (total) volume?

Suppose the base of each partition is x by x and the height is y. The volume is

The area of the bottom is , the area of the front is , the area of the back is , the area of the left side is , the area of the right side is , and the area of the divider that separates the two partitions is . The total area is

Solving for y yields . Plug this into the volume equation and simplify:

x can't be 0, since causes division by 0 in the formula for y. Since x is a length, it must be positive. So , and there is no other restriction on x. Since x is not restricted to a closed interval , I'll use the Second Derivative Test.

Differentiate:

Find the critical points by setting :

Since x must be positive, I get . This gives

Plug into :

Hence, is a local max. Since it's the only critical point, it's an absolute max.

Copyright 2018 by Bruce Ikenaga