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

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.

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 $$\frac{2}{\sqrt{6\pi}}$$