A company wishes to manufacture a box with a volume of 40 cubic feet that is open on top and is twice as long as it is wide.Find the width of the...

Let x and 2x and h be the width ,linth and height of the
box.

Then the volume of the box = length*width*height =
x*2x*h =2x^2h. But 2x^2*h = 40 cubic feet,

Therefore h =
40/2x^2 = 20/x^2.

Now the material of the box whose sides
are x , 2x and h is  proportional to its surface area which is x*2x+2x*h+2(2x)h =
2x^2+6xh = x^2+6*(20/x^2) = x^2+120/x

So we shoild minimum
area  or the minimum  value of x^2+120/x.

Let A(x) =
2x^2+6/x. By calculus,  minimum A(x)  could be  A(c), where c  is a solution of A'(x) =
0 and makes A''(c)  > 0.

A'(x) = 0 gives:
(2x^2+120/x)' = 0. Or

4x -120/x^2 = 0. Multiply by
4.

4x^3 = 120

x = 120/4 =
30

x = 30^(1/3).

Also f"(x)  =
(4x-120/x^2)' = 4-120* (-1/2x^3) = 4+60/x^3 which is > 0 when x=c =
(30)^(1/3)

Therefore the minimum material  is when the the
width = 30^(1/3).