Arectangular box is to have a square base and a volume of 20 ft3. if the material for the base costs $0.28 per square foot, the material for the sides costs $0.10 per square foot, and the material for the top costs $0.22 per square foot, determine the dimensions of the box that can be constructed at minimum cost.

Then the area of the top and bottom are each x^2. 
Then the height of the box is 40/x^2. 
Then each side of the box is 40/x square feet. 
Then the area of all four sides is 160/x square feet. 
Then the total cost is represented by: 
0.30x^2 + 0.05 (160/x) + 0.20x^2 
Expand the middle term and combine x^2 terms: 
0.50x^2 + 8/x 
Rewrite for easy differentiation: 
0.50x^2 + 8x^-1 
Take the derivative: 
x - 8x^-2 
Set the derivative to zero: 
x - (8/x^2) = 0 
Multiply by x^2 
x^3 - 8 = 0 
Add 8: 
x^3 = 8 
Take the cube root: 
x = 2 feet 
So the height is 40/4 = 10 feet 
So the dimensions with minimum cost are: 
2 x 2 x 10 feet.

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The first one is 2/3the second is 1/160the third is 4/125