Old Derivative Problem solved with partial derivatives

Question

Sometimes I get my best understanding when I work older problems with new techniques. So, I wanted to work the old derivative problem of minimizing the surface area of a cylindrical can to hold 1 liter of oil. We solved this in single variable calculus by getting the surface area equation to have only one variable and then taking the derivative and setting it =0. Worked perfectly.

Now, I thought, well, since I know partial derivatives, I'm thinking I don't need to get that Surface area equation to have just one variable. I can leave it as:

$$A= 2\pi rh+ 2 \pi r^2$$

Shouldn't I just be able to find the partial derivative with respect to h and a separate partial derivative with respect to r, set them both =0 and solve to get the critical values.

hmmmmm...when I do this, I get something weird.

(a) $$\frac {\partial A}{\partial r} = 2\pi h + 4\pi r$$

(b) $$\frac {\partial A}{\partial h} = 2\pi r$$

Setting $(a) =0$ I get $2\pi h=-4\pi r$ or $h = -2 r$

HUH? This doesn't make sense. My height is a negative number!

Why doesn't this old calculus problem work using partials?

Answer 1

Here is one way we generalize optimization with multiple variables given a constraint. We start with the following: Surface Area $= 2\pi rh+2\pi r^2$, Volume $= \pi r^2h = 1$

We will now apply the method of Lagrange Multipliers. If you haven't covered this already in your class the Wikipedia link should suffice.

Note that we wish to maximize $f(x)=2\pi rh$ subject to the constraint $g(x) = \pi r^2h-1=0$
We will now construct a function $\mathcal{L}(r,h,\lambda)=f(x)-\lambda g(x)=2\pi rh+2\pi r^2-\lambda(\pi r^2h-1)$ and we will calculate the total gradient

First comes the work finding the partial derivatives

$$\begin{align} &\frac{\partial \mathcal{L}}{\partial r} =\; -2 \pi (h (\lambda r-1)-2 r)\\ &\frac{\partial \mathcal{L}}{\partial h} =\; \pi r (2-\lambda r)\\ &\frac{\partial \mathcal{L}}{\partial \lambda} =\; h \pi r^2-1 \end{align}$$ We now find when the gradient is zero, which occurs when each of the above partial derivatives is zero. $$\nabla_{r,h,\lambda}\mathcal{L}(r,h,\lambda)=0 \iff \begin{cases} -2 \pi (h (\lambda r-1)-2 r) & = 0 \\ \pi r (2-\lambda r) & = 0 \\ h \pi r^2-1 & = 0 \end{cases}$$ We now simplify this a bit, assume the height and radius are non-zero $$\begin{cases} h (\lambda r-1)-2 r & = 0 \\ 2-\lambda r & = 0 \\ h \pi r^2-1 & = 0 \end{cases}$$ From the second equation we get that $\lambda r=2$; substituting this into the first equation gives $$\begin{cases} h-2 r & = 0 \\ h \pi r^2-1 & = 0 \end{cases}$$ This yields the solution $$(r,h) = \left((2\pi)^{-1/3},(4/\pi)^{1/3}\right)$$ This yields a surface area of $3\sqrt[3]{2 \pi}$
Note that this is just so much easier to do the Calculus 1 way. Just note that the condition implies that $$h=\frac{1}{\pi r^2}\implies \text{Surface Area} = \frac{2}{r} + 2\pi r^2$$ Solving this gives the same radius as above, which can be plugged in to the simpler surface area formula to get the answer quickly.