Calculus/calculus

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Question
A soup can is to be produced to hold exactly 128π cubic centimeters of soup. How tall should the can be to minimize the surface area of the can?

thanks!

Answer
Let h be the height of the can and let r be the radius.

The volume is (πr^2)h

The surface area is 4πr + 2πrh ( you can google to find out why this is the surface area of a can )

We know the volume will be 128π  , so (πr^2)h = 128π and then h = 128/r^2


Substitute this expression for h into the expression for surface area.

4πr + 2πrh = 4πr + (2πr)(128/r^2) = 4πr + 256π/r

So we need to minimize surface area S(r) = 4πr + 256π/r where r can be any positive real number.

S'(r) = 4π - 256π/r^2

0 = 4π - 256π/r^2

0 = 1 - 64/r^2

0 = r^2 - 64

0 = (r-8)(r+8)


Using the first derivative test , we see that the function S(r) has an absolute minimum when
r = 8 centimeters.

Since h = 128/r^2 , h = 128/64 = 2


So the surface area is a minimum when the can is 2 centimeters tall.  

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