Question

A rectangular sheet of tin-plate is 2k cm by k cm. Four squares, each with sides x cm, are cut from its corners. The remainder is bent into the shape of an open rectangular container. Find the value of x which will maximize the capacity of the container.

I know the volume is x(2k-2x)(k-2x) but don't know where to go from there

Answer 1

keep in mind that k will be a constant, so ..
V = x(2k^2 - 4kx - 2kx + 4x^2)
= (2k^2)x - (6k)x^2 + 4x^3

dV/dx = 2k^2 - 12kx + 12x^2 = 0 for a max of V
x = (12k ± √(144k^2 - 4(12)(2k^2))/24
= k(12 ± √48)/24
= k(12 ± 4√3)/24
= k(3 ± √3)/8 , where k > 2x

Answer 2

Going from
2k^2 - 12kx + 12x^2 = 0
I get a divisor of 4, not 24

Answer 3

I read the quadratic backwards, for some strange reason!