How to Maximize Box Volume Using Calculus by Maria Clark

I will show you how to solve a common problem found in Calculus classes. The point is to maximize the volume of a box. I will use this example problem to show how this is done:

A rectangular box without lid is to be made from a square cardboard of sides 18 cm by cutting equal squares from each corner and then folding up the sides. Find the length of the side of the square that must be cut off if the volume of the box is to be maximized. What is the maximum volume?

Step 1: Estimate

When you first hear the problem, what is your initial guess as to how large the squares should be? Here, I have done a couple of examples of what the boxes will end up looking like depending on the size of the square you have chosen.

The first image is if you make the squares with 6 cm sides. It forms a cube which is shown in the second image. The third image is if you thought smaller squares would maximize your volume. The squares are formed by 1 cm sides instead. The fourth image is what that box would look like..

Step 2: Find the Equation

In order to maximize the volume, we must first assign a value to the sides of the squares we will be cuting out of the cardboard. We will call that value "x" for now.

Now that we have assigned a value for the sides, we can write the equation of the volume of the box. We must multiply the width by the length by the height. The original piece of cardboard is 18 cm by 18 cm. Therefore after we cut the squares out, each side of the piece of the cardboard will be (18 - 2x). Once we fold up those sides the height will be represented by x.

The volume of the final rectangular box will therefore be: (18-2x)(18-2x)(x).

Step 3: Find the Derivative

In order to find the value of the x that will maximize the volume of the box, we must take the derivative of the equation we found in step 2. After that is found, the equation must be set equal to 0 to get the value of x.

First of all, we have to note that the values of x must be between 0 and 9 because our volume can not be 0 or less then 0. Values in this range will not give us 0 or a negative number.

To take the derivative of the equation, first we must expand it. We must multiply (18-2x)(18-2x)(x) all out. Expanded, this looks like: 4x^2-72x+324.

Next, we take the derivative of this expanded equation. The result of this would be: 12x^2-144x+324. This is our first derivative, and to find x, it is to be set equal to 0.

To make this easier, a 12 can be factored out. 12(x^2-12x+27)=0. Reduced, this becomes 12(x-9)(x-3)=0. After this reduction, we find that x is 9 or x is 3.

Earlier we stated that x must be between 0 and 9. Because of this, the only possible value of x is 3. The value 9 would give us an end volume of 0 cm^3 using the original equation.

Step 4: Use the Second Derivative Test

In order to confirm that x=3 will maximize the volume of this box, we must take the second derivative. If x=3 in the second derivative is a positive number, then the 3 is a critical point giving us a minimum. Therefore it will not maximize our box. If the value is negative, then 3 is a maximum and indeed gives us the maximum value of our box.

The first derivative was 12x^2-144x+324. Taking the derivative of that, in order to get our second derivative, we get: f"(x) = 24x-144.

Plugging in 3 for x, we get: f"(3) = 24(3)-144 = -72.

Step 5: Find the Volume

We now know that x must be 3. Cutting out squares with sides of 3 cm and then folding up the sides will maximize the volume of the rectangular box formed by the piece of cardboard. The pictures shown give a representation of the box with the maximum volume.

In order to get the value of the volume, plug in 3 to the original equation.

(18-2(3))(18-2(3))(3) = 432 cm^3. This is the maximum value.

Step 6: Compare the Results to Estimations

We have solved our problem. Just to prove the point, we can compare the volume of our answer to the volumes of our estimations. Our first estimation was x = 6 cm. Plugging this into the volume equation, we get: (18-2(6))(18-2(6))(6) = 216 cm^3. Our second estimation, with the smaller squares, was x = 1 cm. Plugging this value in, we get: (18-2(1))(18-2(1))(1) = 256 cm^3. Both of these volumes are considerably less than our answer of 432 cm^3.

These volumes can be compared in the above picture with all three of the finished boxes. The box with the 3 cm squares cut out is shown in pink, the box with 6 cm squares cut out is in green, and the box with the 1 cm squares cut out is in blue.

Step 7:

Here is a flow chart for the process just described. This can be helpful to organize your thoughts and is more handy then the whole "how-to" page. This chart is for maximizing or minimizing volumes.