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 (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

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.

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!

To find the dimensions of the largest acceptable box with square ends, we need to consider the given condition that the sum of the length and girth should be at most 108 inches.

Let's assume that the length of the box is x inches.

Since the box has square ends, the width and height will also be x inches.

The girth of the box is the distance around it, which can be calculated by adding the sum of the two equal sides of the box twice.

So, the girth can be calculated as: 2 * (width + height)

Since the width and height are both x inches, the girth is: 2 * (x + x) = 4x

Now, according to the given condition, the sum of the length and girth should be at most 108 inches:

x + 4x ≤ 108

Simplifying the equation:

5x ≤ 108

Dividing both sides of the equation by 5:

x ≤ 21.6

Since the dimensions of a box cannot be a fraction, we need to round down the value of x to the nearest whole number.

Therefore, the maximum value for x is 21.

So, the dimensions of the largest acceptable box with square ends are 21 inches in length, 21 inches in width, and 21 inches in height.