A box has a bottom with one edge 2 times as long as the other. If the box has no top and the volume is fixed at , what dimensions minimize the surface area?

width of base --- x

length of base -- 2x
height --- k, where k is a constant

Volume = c = 2kx^2 , where c is a constant
k = c/(2x^2)

surface area = 2x^2 + 2kx + 4kx
= 2x^2 +4kx
= 2x^2 + 4x(c/(2x^2)
= 2x^2 + 2c/x

d(surface area)/dx = 4x -2c/x^2 = 0 for a min of SA
4x = 2c/x^2
2x^3 = c
x^3 = c/2
x = (c/2)^(1/3)
2x = 2(c/2)^(1/3)
k = c/(2(c/2)^(2/3)
= (1/2)c^(1/3)

To minimize the surface area of the box, we can use the concept of the derivative.

Let's denote the length of the shorter edge as "x". Therefore, the longer edge will be "2x".

The volume of the box is fixed at "V". We can express the volume as the product of the dimensions:

V = x * 2x * h,

where "h" represents the height of the box. Since we don't have a top, the height is not significant in determining the surface area, so we can say that h = V / (x * 2x).

Now, let's express the surface area of the box in terms of x:

Surface Area = 2(x * 2x) + 2(x * h) + 2(2x * h)
= 4x^2 + 2x(V / (x * 2x)) + 4x(V / (x * 2x))
= 4x^2 + V/x + 2V/x + 4V/x
= 4x^2 + 7V/x.

To find the dimensions that minimize the surface area, we need to find the critical points of this equation. We can do this by taking the derivative of the surface area equation with respect to x and setting it equal to 0:

d(Surface Area) / dx = 8x - 7V/x^2 = 0.

Simplifying this equation, we get:

8x = 7V/x^2.

Now, we can solve for x:

8x^3 = 7V.

Therefore,

x^3 = 7V/8.

To find the value of x that minimizes the surface area, we need the positive cube root of the right-hand side of the equation:

x = (7V/8)^(1/3).

By substituting this value of x back into the surface area equation, we can find the corresponding dimensions that minimize the surface area.