Optimization?

Question

a boxwith a square base and open top must have a volume of 32000 cm^3. Find the dimensions of the box that minimize the amount of material used.

Answer 1

let x be the length of the square base, and y be the height.
Given x^2 * y = 32000
Assuming same thickness of the material,
Area of the box, A = x^2 + 4xy = x^2 + 4x * (32000/x^2)
= x^2 + (4*32000)/x

To minimize the material used means to minimize A.
So, find what value is x when dA/dx = 0
2x - (4*32000)/x^2 = 0
x = cuberoot(4*32000/2)
x = 64000^(1/3) = 40
Is this a minimum turning point for A? check d(dA/dx)/dx when x = 40.
d(dA/dx)/dx = 2 + (2*4*32000)/x^3, thus when x = 40, d(dA/dx)/dx > 0, thus at x=40, A is minimum.
Find y, y = 32000/x^2 = 32000/1600 = 20
Therefore, the box's dimension is 20 x 40 x 40.