Prove that the maximum total resistance of two parallel resistors is when the resistors are equal

Question

Taking the two resistors as $x$ and $y$, I have to prove that resistance is maximum when $x = y$. The total resistance is given by $$R = \frac{xy}{x+y}\ $$

How can I solve for the general case, with no relation between x and y, using any branch of mathematics. That is, why is it that for any y, the maximum total resistance will be when x = y

Related but not duplicate: Inequality for the combined resistance of two resistors connected in parallel

Answer 1

$f(x,y) = \frac{xy}{x+y}$, $x\neq -y$

Then we must simultaneously solve $\frac{\partial f}{\partial y} = \frac{\partial f}{\partial x} = 0$ note that these are the partial derivatives of $f$

$\frac{\partial f }{ \partial y} = \frac{(x+y)\cdot x - xy}{(x+y)^2} = \frac{x^2}{(x+y)^2} $

Similarily, $\frac{\partial f}{\partial x} = \frac{y^2}{(x+y)^2}$

Can you proceed from here?

Edit:

Then $$\frac{y^2}{(x+y)^2} =\frac{x^2}{(x+y)^2} = 0$$

$$y^2=x^2=0$$

$\implies y=\pm x$

But we have that $x\neq -y$, so $y=x$

so we're left with the stationary point $x=y=0$, but this implies that $x=-y$, hence there are no stationary points.

It is then not possible to conclude that $x=y$ provides the maximum with the given information.

Answer 2

General case (following EDIT):

The maximum total resistance is infinite, and this is achieved with $x=y=\infty$, i.e. two open circuits (or no circuit at all if you like).

Answer 3

As argued above, if nothing is known about $x$ and $y$, other than that they are positive, there is no definite maximal value. However, the answer suggested may be obtained if it is assumed (not stated in the original problem, though) that the sum $x+y$ is fixed, say, $x+y=\varrho > 0$. Indeed, with this condition, $$R=\frac{x(\varrho-x)}{\varrho} =-\frac{x^2}{\varrho}+x$$ and this quadratic is maximized when $x=-\frac{1}{2\cdot(-\frac{1}{\varrho})}=\tfrac{1}{2}\varrho$, and then also $y=\varrho -x = \tfrac{1}{2}\varrho$.