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?


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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I can answer questions from the standard four semester Calculus sequence. I am not prepared for questions on Tensor Calculus. Everything else is welcome. Derivatives, partial derivatives, ordinary differential equations, single and multiple integrals, change of variable, vector integration (Green`s Theorem, Stokes, and Gauss) and applications.


Ph.D. in Mathematics and many years teaching Calculus at state universities.

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