shafqat said:
Let width be x, length 2x, and the other dimension y. Use the surface area: 4x^2 + 6xy = 300. Solve for y. Then find an expression for the volume by multiplying the 3 dimensions. Maximimum by differentiation.
Finishing this off...
4x^2 + 6xy = 300
6xy = 300 - 4x^2
y = 50/x - (2/3)x
Volume (V) = x * 2x * y = 2(x^2)*y = 100x - (4/3)x^3
V' = 100 - 4x^2
When V' = 0, 4x^2 = 100
x^2 = 25
x = 5 (x is a length, and lengths are positive)
V'' = -8x, which is negative at x = 5.
Therefore x = 5 is an absolute maximum.
Therefore the maximum volume is when dimensions are 5cm, 10cm and 20/3 cm