area of top = x^2
area of sides = 4 x y
area of top = x^2
volume = x^2 y = 20
cost of bottom = 37 x^2
cost of top = 13 x^2
cost of sides = 10 * 4 * x y = 40 x y
total cost = c = 50 x^2 + 40 x y
volume constraint: y = 20/x^2
so
c = 50 x^2 + 40*20/x
dc/dx = 0 for max or min
0 = 100 x - 800/x^2
x = 8/x^2
x^3 = 8
x = 2
then
y = 20/4 = 5
A rectangular box is to have a square base and a volume of 20 ft3. If the material for the base costs 37¢/square foot, the material for the sides costs 10¢/square foot, and the material for the top costs 13¢/square foot, determine the dimensions of the box that can be constructed at minimum cost.
x = ft
y = ft
1 answer