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?