~The Biggest Box~

~This is a classic problem in most Calculus I courses

~I will give you a variation with my own numbers

~At the end of this article I will show you how to solve it using your TI calculator, but the understanding of the basic concepts involved are lost. So, I will be solving it the long way. That way, important concepts are stressed and it would help you apply them to other problems.

~Here is your problem: You would like to make a box from a piece of rectangular cardboard measuring 50" in length and 30" in width by cutting identical size squares from the corners then folding up the sides.

~The problem you have is the size of the squares removed from the corners. Obviously, different size squares will produce different size boxes. Some with very low heights having relatively large bases and others with higher heights having smaller bases. The base consists of the resultant length & width of the box after the cut is made.

~What you want to find is the size of the square cut from the corners that will give you the largest possible box.

~In theory, there are an infinite number of boxes you can make. Some with very little space inside (small volumes) and others with more space (bigger volumes).

~Is there a square cut that produces the largest box (maximum volume)?

~Let's find out. To apply the optimization (max/min) techniques we need to formulate an equation for the quantity to be optimized in terms of the independent variable.

~So, let x represent the side of the square that is cut from each corner. The values of x are obviously restricted between 0 and 15", since it would make no sense out side of this range. Our cardboard is only 30" wide. Of course, a positive value for x is necessary as well.

~After the x" by x" squares are removed from the corners, the length of the resultant box can will be expressed as 50-2x and the width would be represented by 30-2x. The height, when folded upwards would be represented by x.

~So, the volume, V would be represented by the equation, V=(50-2x)(30-2x)x, the product of the three dimensions which would give us all possible volumes for any x in the acceptable range for x.

~At this point in the problem we can take two paths. The derivative method or the calculator method.

~I will show you both ways, but first the derivative method.

~In mathematical terms, we would like to find the value(s) for x between 0 and 15 that maximizes V.

~Critical values for x are found by solving dV/dx = 0. We do not need to look elsewhere since there are no end points for x to check and the equation for V is a polynomial. Polynomials are smooth curves which means that derivatives exist everywhere. So, let's find where dV/dx = 0.

~Differentiating V using the generalized product rule for 3 factors, we get

dV/dx = (-2)(30-2x)x+(-2)(50-2x)x+(1)(50-2x)(30-2x).

~For those of you not familiar with this product rule extension, it goes like this: derivative of first factor times the other factors + the derivative of the second factor times the others + the derivative of the third factor times the others, and so on..

~We need to solve dV/dx=(-2)(30-2x)x+(-2)(50-2x)x+(1)(50-2x)(30-2x) = 0

~To make the numbers smaller, divide both sides by 4, we get

(-1)(15-x)x+(-1)(25-x)x+(1)(25-x)(15-x)=0. Of course, 0 divided by 4 = 0

~Eliminating all symbols of inclusion, we get,
-15x+x²-25x+x²+375-40x+x²=0 or,
3x²-80x+375=0, a quadratic equation. Let's see if it factors by computing the discriminant, b²-4ac. We have a=3, b=-80, and c=375, so b²-4ac=1900 (not a perfect square)

~So, we need to find the solutions using the quadratic formula

~So, x=[-(-80)+√1900]/2(3) and x=[-(-80)-√1900]/2(3)

~Which gives, x=20.60 and x=6.07 to two decimal places. We can reject x=20.60 since it is outside of the possible values for x. So, x=6.07" seems to be the answer. How can we be sure?

~We can use our common sense and conclude that it must give a maximum volume since there is no minimum. (i.e., taking x very close to zero in the formula for V will make V as small as we want.

~The other way is to graph V and locate the maximum by using our TI calculator. Let me show you this procedure.

~First, code in the equation for V in the y= menu (upper left). Make sure you code in V not dV/dx.

~Go to the table of values for V (2nd Graph). Make sure your table is set up properly so you can enter various x values.
If not, go to 2nd window (table set) and see if the Indpnt is on ASK and the Depend is on AUTO.

~Now enter a few values between 0 and 15 & observe the V values. Notice that no V value is greater than 5000. Now, set up your window with x_min=0. x_max=15, y_min=0, y_max=5000

~Now, press graph. Hopefully, you should be viewing the Volume curve and also observing that the curve reaches a max point somewhere

~We hope that the x value at that max point is consistent with what we got earlier, namely x=6.07

~Let's find out. Press 2nd trace (calc) and go to menu 4 (Maximum), enter. Now, you will be prompted to give a guess for x to the left of the high point. If you are not sure where you are, use your courser to move the marker along the curve & see those x values. I entered x=4 as a left guess. Any x to the left of the high point will work. Then press enter. Now, you will be prompted for a right guess. I entered x=8. Again, any guess to the right of the high point will work. Press enter. Now, enter any guess between your left guess & your right guess. I entered x=7. Press enter and the solution for x should appear at the bottom of the window, Yippie!, x=6.07.

~Therefore, you would cut 6.07" by 6.07" squares from the corners of your cardboard to get the box (with an open top) that holds the most. (i.e., Maximum Volume)

~I hope you enjoyed this problem