# A Different Approach to Optimization Problems

Historically, maximization problems have been a problem for calculus students. Not so much in finding critical points from functions and classifying them, but finding the objective function to begin with. Students rely on basic examples to mimic. We wonder why they can’t carry out more complicated examples. They are able to model simple physical situations after much practice, but flounder when faced with slightly more complicated problems because they don’t understand the complication well.

As the maximization and minimization problems become more complicated, using a table to organize examples of what is being modeled becomes more and more useful. Let’s look at another examples of how this might work.

Problem A rectangular tank with a square base, an open top, and a volume of 8788 ft3 is to be constructed of sheet metal. Find the dimensions of the tank that has the minimum surface area.

Let’s create a table with various size containers and their corresponding surface areas. We can use this table to find the objective function, the function to be minimized.

Let’s start with a container with a very small base, 1 foot by 1 foot. To have a volume of 8788, the height of the container must be 8788 ft (Volume = length width height)

side area = 4 1 8788

This gives a total area of

total area = 1 1 + 4 1 8788 = 35153

Let’s make this the first row of the table.

Let’s make the base of the container bigger. Instead of a 1 ft by 1 ft base, make the base 10 ft by 10 ft.

This container has a bigger base, but is not as high. The height must be 87.88 ft so that the volume is 8788 ft3 (Volume = 10 10 87.88). The base area is

base area = 10 10

and the area of the sides is

side area = 4 10 87.88

This gives a total area of

total area = 10 10 + 4 10 87.88 = 3615.2

and make the second row of the table.

From these two examples, we get a good idea of how the dimensions relate to the surface area of the container. Applying this same reasoning, we can put another example in the table.

Let’s put all of this together in a larger table.

Examine the table. Notice how each dimension contributes to the base area and the side area. As the base gets larger, the total surface area decreases. However, is there a point at which making the base larger causes the total surface area to increase?

To answer this question, let’s generalize the dimensions. If the base is x feet by x feet, the height must be 8788/x2 so that the volume is

Volume = x x 8788/x2 = 8788

The base area is

base area = x x

and the area of the sides is

side area = 4 x 8788/x2

This gives a total area of

total area = x x + 4 x 8788/x2

This leads the last row in the table.

Based on this last row, define the total area function A(x),

where the function has been simplified to make it easier to work with. The critical points of this function are more easily found if we rewrite the function as

The derivative is

The critical point is found by setting the derivative equal to zero. It is easier to solve the resulting equation if rewrite all exponents as positive,

We need to determine whether the critical point is a relative maximum or a relative minimum. We could use the first derivative test, but it is easier to test the critical point for concavity using the second derivative. The second derivative is

The sign of the second derivative at x = 26 is positive,

so the function is concave up at the critical point. This makes the critical point a relative minimum. The dimensions of the box are

x by x by or 26 ft by 26 ft by 13 ft

Give this a try. Then have your students attempt a similar problem…instead of a square base, try a base in which the length is twice the width. My student were easily able to modify their strategy using a table. This strategy also works well for economic lot size and economic order quantity problems too.