A box is open and its length is twice its width

a box is open and its length is twice its width. If 15000 cm of steel is used in making it, what is the maximum volume and what is its height?

Standard equation for volume of the box is V = L*W*H
W = width of the box
L = length of the box = 2*W
H = height of the box
V = (2*W)*W*H = 2*W^2*H

Determine the relation for the surface area of the box.
The box is open so there is no top.
A1 = area of bottom = L*W = 2*W*W = 2*W^2
A2 = each end = W*H
A3 = each side = L*H = 2*W*H
Area = A1 + 2*A2 + 2*A3 = 2*W^2 + 2*W*H + 4*W*H = 2*W^2 + 6*W*H = 15000

Solve this for H in terms of W:
6*W*H = 15000 - 2*W^2
H = (15000 - 2*W^2)/(6*W) = (7500 - W^2)/(3*W)

Use this in the equation for the volume.
V = 2*W^2*H = 2*W^2*[(7500 - W^2)/(3*W)]
V = (2/3)*W*[7500 - W^2] = (2/3)*[7500*W - W^3]

Find dV/dW and set it equal to 0 to get the maximum.
dV/dW = (2/3)*[7500 - 3*W^2] = 0
7500 - 3*W^2 = 0
2500 - W^2 = 0
W = sqrt(2500) = 50
So the width for maximum volume is 50
Use this in the equation for volume to get the maximum volume.

Maximum volume = (2/3)*[7500*W - W^3] = (2*W/3)*[7500 - W^2]
V = (2*50/3)*5000 = 500000/3 = 166666.6667 cm^3

Height = (7500 - W^2)/(3*W) = 33.3333 cm