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#