Posted by **Ben** on Monday, July 18, 2011 at 1:54pm.

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.

- calculus -
**Reiny**, Monday, July 18, 2011 at 2:11pm
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.

- calculus -
**Ben**, Wednesday, July 20, 2011 at 4:42pm
Why did the 2ðx^2+108/x^2 become 2ðx-108/x^2

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