Question

A parcel delivery service a package only of the length plus girth (distance around) does not exceed 24 inches.

A) Find the dimensions of a rectangular box with square ends that satisfies the delivery service's restriction and has a maximum volume. What is the maximum volume?

B) Find the dimensions (radius and height) of a cylindrical container that meets the delivery service's restriction and has a maximum volume. What is the maximum volume?

Answer 1

Let the size of the square (cross-section) be s.
Then we need to maximize
V=s²(24-2s) with respect to s.
First find the derivative and equate to zero:
dV/ds = 48s-6s²=0
means
s=0 or s=8
s=0 corresponds to a minimum volume and
s=8 corresponds to a maximum volume.
So the maximum volume is given by
8x8x8 inches, as is evident by intuition.

Answer 2

but an 8x8x8 box has length+girth = 8+32 = 40 inches, so it will not work.

We need to optimize s^2(24-4s) since a square has 4 sides.

v = 24s^2 - 4s^3
v' = 48s - 12s^2
v'=0 when s=4

So, a 4x4x8 box has max volume.

Do (B) similarly

Answer 3

Good catch, Steve.
Thank you.