you have a rectangular piece of cardboard 40X48 inches and need to create an open box with the maximum volume(ignore flaps). what would the dimensions of the box need to be to maximize the volume?

Answer 1

v = x(40-2x)(48-2x) = 4x(20-x)(24-x)
v=0 at x=0,20,24, so you know there will be a max for 0 < x < 20

If you haven't had some calculus, this will be hard to find; you may have to do some probing around x=10 and try some values.

If calculus is in your bag of tools, then you know that

dv/dx = 4(x^2 - 88x + 480)
dv/dx=0 when x = 4/3 (11-√31) = 7.24

So, the box is 32.76 x 40.76 x 7.24