Minimizing The Cost

I have this exercise that I would like anyone to suggest the required steps in order to solve it

A cylindrical can is to be made to hold $250 \pi\; cm^3$. Find the dimensions of the can that will minimize the cost of the metal to manufacture it.

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Volume is

$$V=\pi r^2 h = 250\pi \implies h = \frac{250}{r^2}$$

And thus, the surface area is given by

$$A = 2\pi r^2+2\pi r h =\\ 2 \pi r^2 + 2\pi r\frac{250}{r^2} =\\ 2 \pi r^2 + 2\pi \frac{250}{r} =\\ 2 \pi r^2 + \frac{500 \pi}{r}$$

Now minimize $A$.

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Why isn't the r in the denominator squared? – aablah Dec 2 '12 at 9:25
@aablah $$2 \pi r h = 2 \pi r \frac{250}{r^2} = 2 \pi \frac{250}{r}$$ – Argon Dec 2 '12 at 15:53

Steps:

• First note that the volume is $V(r,h) = \pi r^2h = 250\pi$. That is $r^2h = 250$.
• Now the surface area is $A(r,h) = 2\pi r h + 2\pi r^2$. The term $2\pi r^2$ is the area of the top and bottom.
• You want to minimize the surface area, so find how the area depends on for example the radius alone. You by taking the first equation and finding how $h$ depends on $r$: $h = 250 r^{-2}$. Put that into the formula for $A$. This eliminates the variable $h$ and you have the area as a function of the one variable $r$.
• Now you minimize this function (which is now of one variable: $A(r)$) by taking the derivative and setting it equal to zero. You solve for $r$.
• You find the critical point (a value of $r$) and check that this indeed corresponds to a minimum.
• Then you can use the previous formula to find the value of $h$.
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what about the top and bottom of the cans surface area :) – MSEoris Dec 1 '12 at 20:26
@MSEoris: oops :) – Thomas Dec 1 '12 at 20:26