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 example of how this might work.

**Problem** A rectangular container with a square base, an open top, and a volume of 8788 ft^{3} 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 vary 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). To find the surface area, break the area into the area of the base and the area of the four sides. The base area is

base area = 1 · 1

and the area of the sides is

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/*x*² 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 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