The U.S. Post Office will accept a box for shipment only if the sum of the length and girth (distance around) is at most 108 inches. Find the dimensions of the largest acceptable box with square ends.

Answer 1

v = vol = x^2 y
girth = 4 x
length = y
so y + 4x </=108
since maximizing
y + 4x = 108
or
y = 108 - 4x

v = x^2 (108-4x)
v = 108 x^2 - 4 x^3
dv/dx = 216 x - 12 x^2
= 0 for max or min
so
x(216 - 12x) = 0
x = 18 for max
then y = 108 -4(18) = 36

Answer 2

Let the width and the height of the box both be x inches. (That makes the end a square)
Let the length by y inches
distance around lengthwise = 2x + 2y
distance around widthwise = 4x
(think of a ribbon around a box when you wrap for Christmas)
distance = 6x+2y= 108
y = 54 - 3x

I will assume that by "largest box" you mean greatest volume.
V = x^2 y
= x^2(54-3x)
= 54x^2 - 3x^3
dV/dx = 108x - 9x^2 = 0 for a max V
9x^2 = 108x
x = 12
then y = 54 - 36 = 18
the largest box is 18" long, 12" wide, and 12" high.

Answer 3

Reading the question again, I think that I took the wrong interpretation and Damon took the right one.

Answer 4

It is just length + x^2
not girth in both directions
http://pe.usps.com/text/qsg300/Q401.htm

Answer 5

I mean length + 4x

Answer 6

Both of you, thank you very much!!! I arrived at the correct answer width = 18 and length = 36, but I just got that answer by chance and wasn't sure how I could prove (mathematically) that it was indeed correct, your explanations helped tremendously!