calculus
posted by Ben .
A large soup can is to be designed so that the can will hold 54pi cubic inches of soup. The radius of the can is "x" inches and the height of the can is "h" inches. Find the values for the radius and the height such that the amount of metal needed is as small as possible.
Note: The side require 2pixh square inches of metal and each end requires pix^2 square inches.

Volume = πx^2h
h = 54/(πx^2)
Surface area = SA = 2πx^2 + 2πxh
SA = 2πx^2 + 2πx(54/(πx^2))
= 2πx^2 + 108/x
d(SA)/dx = 2πx  108/x^2
= 0 for a min of SA
2πx = 108/x^2
x^3 = 108/(2π) = 54/π
x = (54/π)^(1/3) = appr. 2.58076
h = 54/(πª2.58076^2)) = appr. 2.58076
surprise! , the height should be equal to the radius. 
Why did the 2ðx^2+108/x^2 become 2ðx108/x^2
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