Calculus optimization: dimensions of a box

Find the dimensions of the base of the rectangular box of greatest volume that can be constructed from 200 square inches of cardboard if the base is to be three times as long as it is wide?

Suppose that the length of the box is x. Since the base is three times as wide as it is long, the width is 3x. If we call the height y, then the surface area of the box in terms of x and y is:
S = 2LW + 2LH + 2WH
= 2(x)(3x) + 2(x)(y) + 2(3x)(y)
= 6x^2 + 8xy.

Since we have 200 in^2 of cardboard:
6x^2 + 8xy = 200 ==> 3x^2 + 4xy = 100.

The volume of the box is:
V = LWH
= (3x)(x)(y)
= 3x^2*y,

so we want to maximize V = 3x^2*y with respect to 3x^2 + 4xy = 100. If we solve 3x^2 + 4xy = 100 in terms of y, we can get V in terms of x.

Solving 3x^2 + 4xy = 100 for y gives:
y = (100 - 3x^2)/(4x).

Then, the volume of the box in terms of x is:
V = 3x^2*y
= 3x^2(100 - 3x^2)/(4x)
= 3x(100 - 3x^2)/4
= (300x - 9x^3)/4.

Differentiating this with respect to x gives:
dV/dx = (300 - 27x^2)/4.

Setting dV/dx = 0 yields:
300 - 27x^2 = 0 ==> x = 10/3.

Therefore, the width of the base is 10/3 inches and the length of the base is 3(10/3) = 10 inches.

I hope this helps!

200=3L x L
200= 4L
50= L

so the length is 150, the width is 50