```Question 112462
Let's say you have a square piece of sheet aluminum that is 12 inches on each side.  You want to cut a square of dimension x by x from each corner of the sheet and then fold the sides up to make an open topped box.  What dimension should you select for x so that box you make has the maximum possible volume?

If the original sheet of metal is 12" by 12", and you cut an x by x square from each corner, the dimensions of the bottom of the box would then be 12 - 2x inches.  Then the volume with respect to the x dimension of the box would be {{{V(x)=(12-2x)(12-2x)x}}}.

First, let's test a couple of values to see if it makes any difference.  Let's try x = 1.  x = 1 means that our box would be 10 by 10 on the bottom and 1 inch tall for a total volume of 100 cubic inches.  Let's try x = 3.  x = 3 means that the box would be 6 by 6 on the bottom and 3 inches tall, 6 X 6 X 3 = 108 cubic inches.  So clearly the selected value for x makes a difference in the volume.

{{{V(x)=4x^3-48x^2+144x}}} is the polynomial function, and the real life problem would be to find the value of x that makes the volume a maximum.

{{{graph(400,400,-1,12,-20,150,4x^3-48x^2+144x)}}}

The graph illustrates the situation.  Values for x larger than 6 don't have any application because if x were 6 or larger, there wouldn't be any metal left to make a box.  So the interval of interest is {{{0<x<6}}}

Ok.  Extra credit.  It looks like the x value at the maximum is about 2.   What is the volume of the box if x = 2?```