The corners are removed from a sheet of paper that is `3` ft square. The sides are folded up to form an open square box. What is the maximum volume of the box?

Let us set up the following variables:

`{(x, "Height of box (ft)"), (y, "Width of box (ft)"), (V, "Volume of the box ("ft^3")") :}`

The box has an open top and is square, so it consists of a base, and four identical sides.

Then the width (and therefore length) of the box is given by:

`x + y + x = 3 => y=3-2x`

Note that one really important constraint on `x` and `y` is that:

`x gt 0`
`y gt 0 => 3-2x gt 0 => x lt 3/2`

Thus we must have `0 lt x lt 3/2`, Without this constraint the box cannot physically be constructed! The Volume of the box is then:

`V =`(area of base) `xx` (height)
`\ \ = y * y * x`
`\ \ = xy^2`
`\ \ = x(3-2x)^2`
`\ \ = 4x^3-12x^2+9x`

If we graph this volume function we get `("remember " 0 lt x lt 3/2)`
graph{4x^3-12x^2+9x [-1, 3, -1.74, 3.26]}

It appears from the graph that we have a maximum volume of `2 ft^2` when `x=1/2`, and a minimum volume of `0` when `x=3/2`, so let us examine this further:

At a critical point (max or min) the derivative, `(dV)/dx` will vanish. Differentiating wrt `x` we get:

`(dV)/dx = 12x^2-24x+9`

At a critical point we have:

`(dV)/dx = 0 => 12x^2-24x+9 = 0`
`:. 4x^2-6x+3 = 0`
`:. (2x-1)(2x-3) = 0`

Leading to two possible solutions:

`x =1/2, 3/2`

This is completely consistent with our graph so now let us prove the nature of the turning points by looking at the second derivative:

`(d^2V)/(dx^2) = 24x-24`

So:

`x = 1/2 => (d^2V)/(dx^2) < 0 =>` maximum
`x = 3/2 => (d^2V)/(dx^2) > 0 =>` minimum

Ths we get a maximum volume when:

`x = 1/2 => V=(1/2)(3-2(1/2))^2 = 2`

Which matches are graphical observations.

Thus the maximum occurs when:

`x = 1/2 => y = 2`

Leading to:

`V=2 \ ft^3`