A sheet of paper 60

Question

A sheet of paper 60 cm-by-68 cm is made into an open box (i.e. there's no top), by cutting x-cm squares out of each corner and folding up the sides. Find the value of x that maximizes the volume of the box. Give your answer in the simplified radical form.

Answer 1

V = x(60 - 2x)(68 - 2x)

Can you expand, differentiate, and answer?

Answer 2

First draw a rectangle of length 68 and width 60. Next, cut out squares of length x from each of the corners. When you fold up the flaps you'll have a box of length (68-2x), width (60-2x) and height (x). The volume is then

V = (68-2x)*(60-2x)*x

V = 4x³ - 256x² + 4080x

Note that in order to have a proper rectangle each of the height, length, and width must be greater than 0:

68-2x > 0, 60-2x > 0, x > 0

This simplifies to 0 < x < 30, which is the domain for x

Next, take the derivative of V and set it equal to 0:

dV/dx = 12x² - 512x + 4080 = 0

x = (512 ± √(512² - 4(12)(4080)) / 24

x = (512 ± √(66304)) / 24

x = (512 ± 16√(259)) / 24

x = (64 ± 2√(259)) / 3

x = (64 - 2√(259)) / 3 or x = (64 + 2√(259)) / 3

The value x = (64 + 2√(259)) / 3 = 32.1 is not in the domain for x, thus x = (64 - 2√(259)) / 3 = 10.6

We must finally check to make sure this value is indeed a maximum for the volume. We can use the second derivative test to check. If the second derivative is negative, then the value for x is a local maximum.

d²V/dx² = 24x - 512

Plug in x = (64 - 2√(259)) / 3

24((64 - 2√(259)) / 3)-512 = -257.5 < 0

Thus x = (64 - 2√(259)) / 3 is the value for x that maximizes the volume of the box