Calculus (Optimization)
posted by Mishaka .
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.

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 (1084x)
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 
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(543x)
= 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. 
Reading the question again, I think that I took the wrong interpretation and Damon took the right one.

It is just length + x^2
not girth in both directions
http://pe.usps.com/text/qsg300/Q401.htm 
I mean length + 4x

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!
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