Page 102 - Calculus Demystified

CHAPTER 3 Applications of the Derivative
the sheet and then folding up the sides (see Fig. 3.12). What is the box of 89
greatest volume that can be constructed in this fashion?

Fig. 3.12

SOLUTION
It is important in a problem of this kind to introduce a variable. Let x be the
side length of the squares that are to be cut from the sheet of cardboard. Then
the side length of the resulting box will be 12 − 2x (see Fig. 3.13). Also the
height of the box will be x. As a result, the volume of the box will be
3
2
V(x) = x · (12 − 2x) · (12 − 2x) = 144x − 48x + 4x .
Our job is to maximize this function V .

x

_
12 2x
Fig. 3.13
2

Now V (x) = 144 − 96x + 12x . We may solve the quadratic equation
2
144 − 96x + 12x = 0
to find the critical points for this problem. Using the quadratic formula, we

find that x = 2 and x = 6 are the critical points. Now V (x) =−96 + 24x.
Since V (2) =−48 < 0, we conclude that x = 2 is a local maximum for the

problem. In fact we can sketch a graph of V(x) using ideas from calculus and
see that x = 2 is a global maximum.

We conclude that if squares of side 2 are cut from the sheet of cardboard
then a box of maximum volume will result.

Observe in passing that if squares of side 6 are cut from the sheet then (there
will be no cardboard left!) the resulting box will have zero volume. This value
for x gives a minimum for the problem.