The base of a rectangular box is to be twice as long as it is wide. The volume of the box is 256 cubic inches. The material for the top costs $0.10 per square inch and the material for the sides and bottom costs $0.05 per square inch. Find the dimensions that will make the cost a minimum.

Answer 1

width = w
length = 2w
height = h
w*2w*h = 256, so h = 128/w^2
c = 0.10*w*2w + 0.05(2w^2 + 2h(w+2w))
= 0.2w^2 + 0.5(2w^2 + 2(128/w^2)*3w))
= 1.2w^2 + 384/w
dc/dw = 2.4w - 384/w^2
So, where does dc/dw = 0?

Answer 2

side = 2 x
end = 1 x
height = h
volume = 2 x^2 h = 256
so x^2 h = 128
area top = 2 x^2
cost top = .1*2 x^2 = 0.2 x^2
area sides = 2 * 2x * h = 4 x h
area ends = 2 * x * h = 2 x h
so cost sides and ends = 0.05 (6 x h)= 0.3 x h
cost bottom = 0.05 x *2x = 0.1 x^2
total cost = 0.3 x^2 + 0.3 x h
but as we said x^2 h = 128 so h = 128/x^2
so
cost = 0.3 [ x^2 + x (128/x^2) ]
c = 0.3 ( x^2 + 128/x)
take derivative
dc/dx = 0.3 (2 x - 128/x^2)
that is zeero when
2 x = 128/x^2
x^3 = 64
x = 4