7. A person is going to make a box by taking a square piece of cardboard, which is 12 inches on a side, like the one in the picture to the right, cutting squares out of the corners, and folding the edges like the other figure in the picture to the right. How big of a square should they cut out of the corners in order to maximize the volume of the resulting box?
This is a classical problem which has delighted calculus students down through the ages.
We first express the volume of the box as a function of x, the side of the square which is cut out of each corner.
The resulting box will have a base which is a square which all of whose sides will have length 12 - 2x. The height of the box will be x. Therefore, the volume will be
Since the amount you cut out can't be negative, and since you can't cut out more than half of the side, the domain of the function is
If we differentiate the volume and set it equal to zero, we get
This gives us a quadratic to solve. Factor
Set the factors = 0.
This gives us two points to check
This is probably not the maximum volume. Let's try x = 2
We knkow that the max's and the min's happen either at the end points of the domain or when the first derivative is equal to zero or does't exist. Since the first derivative always exists for a polynomial, we need only worry about the end points of the domain and places where the first derivative is zero. We have the points where the first derivative is zero, and one of them happens to be an end point of the domain. We need only check the other end point of the domain
We conclude that the minimum possible volume is 0 which happens at the end points, and the maximum possible volume is 128 when x = 2.
The two situations at the endpoints of the domain are degenerate cases. When x = 0, you just have your original sheet of cardboard from which you have not yet cut out the corners let alone fold up the edges. If x = 6, the four 6x6 square which you have cut out will use up all of the 12x12 sheet of cardboard leaving you with nothing with which to make your box. If you were to argue that these degenerate cases should not be considered, then there is no minimum volume. While you can't get to x = 0 or x = 6, the closer you come to them the smaller the volume. So no matter how small of a volume you have for your box, you could simply take an x which is closer to the endpoint, and you would get a smaller volume. This illustrates one reason why it is important to be working over closed intervals when doing max and min problems.
The graph looks like
You can see that the volume peaks out at around 128 when x = 2.
When we look at the graph, we see that the formula for the volume doesn't know that x has to be between 0 and 6. If x is less than 0, the factor of the height is negative, but the 12 - 2x becomes even larger when we subtract negative numbers with larger absolute values. The result is negative volume. Since one could argue that you can't have negative volume, we would in that case have to reject negative values for x. However, if x > 6, we see that the volume becomes positive again. That is because the negative factor, 12 - 2x, is squared, and squaring wipes out negatives.