If the dimensions are x,z, with z being the height, then
x^2 z = 78, so z=78/x^2
The cost c is
c(x,z) = .30x^2 + 4(.10xz) + .20x^2
But, z=78/x^2, so
c(x) = .30x^2 + .4x(78/x^2) + .20x^2
c(x) = .50x^2 + 31.2/x
Now that's not easy to do algebraically, but with a little calculus we have
dc/dx = x - 31.2/x^2 = (x^3-31.2)/x^2
dc/dx=0 when x^3-31.2=0, or
x=3.148
z=78/3.148^2 = 7.870
So, the box has is 3.148x3.148x7.870
A rectangular box is to have a square base and volume of 78 feet cubed. If the material for the base costs $.30 per square feet, the material for the sides costs $.10 per square foot, and the material for the top costs $.20 per square foot. Determine the dimensions of the box that can be constructed at minimum cost.
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1 answer