Math question please optimization?

Question

US postal service will accept a box for shipment only if the sum of its length and girth (the distance around the box) does not exceed 108 inches. What dimensions will give a box with a square end the largest possible volume..

IDK how to solve it but the answers are 18 x 18 x 36

Answer 1

Represent the sides with variables. For example, use s for the length of the side of the square, and L for the length.

4s + L <= 108
L <= 108 - 4s
Obviously the greatest volume will occur with the greatest dimensions, so L = 108 - 4s

Represent the volume, substituting this specific length information:
V = L*s*s
V = (108-4s) * s * s
V = 108s^2 - 4s^3

To get maximum, take derivative and find where it is zero:
V ' = 216s - 12s^2
216s - 12s^2 = 0
s(216 - 12s) = 0
s = 0 or
12s = 216
s = 18

(should have checked to see if this is maximum or minimum also.)
So the side of the square should be 18 inches.

Answer 2

Let
L = length
w = width
h = height
S = sum of dimensions
V = volume

Given

w = h
S = L + 2w + 2h = 108

We have:

S = L + 2w + 2h = L + 2w + 2w = L + 4w = 108
V = Lwh = Lw²

Solve for L.

S = L + 4w
L = S - 4w

Substitute into formula for volume.

V = Lw² = (S - 4w)w² = Sw² - 4w³

Take the derivative with respect to w and set equal to zero to find critical points.

dV/dw = 2Sw - 12w² = 0
Sw - 6w² = 0
w(S - 6w) = 0
w = 0, S/6

Take the second derivative to determine the nature of the critical point.

d²V/dw² = 2S - 24w

For w = 0
2S - 24w = 2S > 0
Implies relative minimum. Therefore solution is rejected.

For w = S/6
2S - 24(S/6) = 2S - 4S = -2S < 0
Implies relative maximum which is what we want.

w = S/6 is the only solution.

Solve for h and L.

h = w = S/6 = 108/6 = 18

L = S - 4w = S - 4(S/6) = S - 2S/3 = S/3 = 108/3 = 36

So the dimensions of the box with maximum volume are

18 x 18 x 36

Answer 3

Let hte dimensions be x,x,y
So 2(y + x) =108
y = 54 - x

V = x^2 * y
= x^2 * (54-x)
= 54x^2 - x^3

dV/dx = 108x - 3x^2
This is zero when x = 36
y = 54 - 36 = 18