Expert: Scotto - 4/1/2009

Question
An open box is to be made from a rectangular piece of material by cutting equal
squares from each corner and turning up the sides. Find the dimensions of the
box of maximum volume if the material has dimensions 6in x 6in

Answer
In this problem, the height of the box is x.  Since this needs to be cut out of all corners, it takes 2x from the width and 2x from the depth.  The volume is then x(6-2x)(6-2x).

To find the value of x that maximizes this function, we need to take the derivative.  Now we could use the product rule for a product of three functions (f'gh + fg'h fgh'), but it would be easier to multiply f, g, and h together first.

We know that 6-2x squared is 36 - 24x + 4x².
If we multiply by another x we get V(x) = 36x - 24x² + 4x^3.
I introduced V(x) as the volume.

We can see that what V'(x) is.
It is V'(x) = 36 - 48x + 12x².

This is easily divisible by 12, giving V'(x) = 12(3 - 4x + x²).

We know that 3 is 1*3, and 1+3 is 4, so the factors involve a -1 and -3 since the middle is -4.  That would mean it factors to
12(3-x)(1-x) { or 12(x-1)(x-3); either one is the same }.

That would mean the values to look at would be x=1 and x=3.

If we only cut x=1, we would have a volume of 1(4)(4).  That is 16.

If we cut x=3, the volume would be 3(0)(0) = 0.  That's must be the minimum.

This means the first would be the maximum.  That's how cubic equations look - they have one minimum and one maximum.

Now just to make sure, lets take the second derivative from
V'(x) = 12(3 - 4x + x²).  We get V"(x) = 12(-4 + 2x).
At x=1, -4+2x is -2.  Times 12 this is -24.  When the second
derivative is negative, that implies the value at that point
is a maximum.