Arectangular box with a volume of 500 ftcubed is to be constructed with a square base and top. the cost per square foot for the bottom is 15cents, for the top is 10cents, and for the sides is 2.5cents. what dimensions will minimize the cost?

The volume is given by:
 V = hx² = 500
 We can then clear the height in terms of x:
 h = 500 / x².
 Area of the base: x²
 Areal from the top: x²
 Area of the vertical walls: 4xh = 4x (500 / x²) or 2000 / x
 The formula sought will then be:
 C = 0.15 (area of base) + 0.10 (area of top) + 0.025 (area of vertical walls)
 C = 0.15x² + 0.10x² + 0.025 (2000 / x)
 Rewriting:
 C = .25x² + 50 / x
 We derive and match zero to find the critical point:
 0 = 0.5x - 50 / x ^ 2
 Clearing x we have:
 50 / x ^ 2 = 0.5x
 x ^ 3 = 50 / 0.5
 x ^ 3 = 100
 x = 4.64
 We derive again:
 C '' (x) = 0.5 + 100 / x ^ 3
 We evaluate x = 4.64:
 C "(4.64) = 0.5 + 100 / (4.64) ^ 3
 C '' (4.64)> 0 (x = 4.64 is a minimum)
 Solving
 h = 500 / x²
 h = 500 / (4.64) ²
 gives
 h = 23.22.
 
 The dimensions of the box are 4.64 x 4.64 x 23.22.
 The total cost would be $ 16.16.

Magic man is here, but unfortunately he is broke won’t you him out