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You are to manufacture a tin can (a closed cylinder) with a total surface area of one square foot. What dimensions do you make the can so that the volume is maximized?

I know the answer is $radius = 1/\sqrt{6π}$ and $height = 2/\sqrt{6π}$ or $\sqrt{6π}/3$.

What I don't know is how to get that answer. I've been trying to use the formulas for the surface area of a cylinder ($2πrh+2πr^2$) and the volume of a cylinder ($πr^2h$), but I can't seem to figure it out. I've been taking the derivative, setting it to zero, and solving for $x$, but no luck.

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The two given answer for $h$ are not equivalent. I will assume the first is correct as that is what I got when solving the problem for myself. I believe the second one is supposed to be $h=\frac{\sqrt{6\pi}}{3\pi}$ There cannot be two distinct possible values for $h$ because the second gives a surface area greater than 1.

First, you need to express $h$ in terms of $r$ this can be done by setting the formula for surface area equal to one and solving for $h$

$$1=2{\pi}rh + 2{\pi}r^2$$ $$1-2{\pi}r^2= 2{\pi}rh $$ $$h=\frac{2{\pi}r^2}{2{\pi}r}$$

Then, you substitute $h$ for that expression in the function for volume and simplify.

$$V(r)={\pi}r^{2}\frac{1-2{\pi}r^2}{2{\pi}r}$$ $$V(r)=\frac{r-2{\pi}r^3}{2}$$

Next, compute $\frac{dV}{dr}$ and set it equal to zero. I throw away the $\frac{1}{2}$ because it won’t matter when we set the derivative to zero. This derivative isn’t too complicated, we just apply the power rule each term in $V(r)$

$$V(r)=\frac{1}{2}(r-2{\pi}r^3)$$ $$\frac{dV}{dr}=\frac{1}{2}(1-6{\pi}r^2)$$ $$0=1-6{\pi}r^2$$ $$r=\frac{1}{\sqrt{6{\pi}}}$$

Finally, plug $r= \frac{1}{\sqrt{6{\pi}}}$ into our equation for $h$

$$h=\frac{1-2{\pi}(\frac{1}{\sqrt{6{\pi}}})^2}{2{\pi}\frac{1}{\sqrt{6{\pi}}}}$$

This simplifies to

$$h=\frac{\sqrt{6{\pi}}}{3\pi}$$

Which also equals $\frac{2}{\sqrt{6\pi}}$

And those are your two answers.

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