Find the $\max$ and the $\min$ of $f(x,y)=xy$ subject to $g(x,y)=\frac{x^2}{a^2}+\frac{y^2}{b^2}-1=0$, using Lagrange multipliers

Question

Find the $\max$ and the $\min$ of $$f(x,y)=xy$$ subject to the constraint $$g(x,y)=\frac{x^2}{a^2}+\frac{y^2}{b^2}-1=0$$ using Lagrange multipliers.

My Attempt:

Answer 1

So if you know $x = \dfrac{-ya^2}{2\lambda}$, then $$ y = x \frac{-b^2}{2\lambda} = \frac{-ya^2}{2\lambda} \times \frac{-b^2}{2\lambda}, $$ so either $x = y=0$ (which is impossible) or $a^2b^2 = (2\lambda)^2 \iff \lambda = \pm ab/2$.

You can similarly plug into the equation of the circle, just plug only for one of$x,y$ so you come out with expression in one vairable.

Answer 2

As you wrote, you want to solve the system$$\left\{\begin{array}{l}y+\frac{2\lambda}{a^2}x=0\\x+\frac{2\lambda}{b^2}y=0\\\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.\end{array}\right.$$So, $x$ is both equal to $-\frac{2\lambda}{b^2}y$ and to $-\frac{a^2}{2\lambda}y$. So, since there is no solution with $x=y=0$, you must have $-\frac{2\lambda}{b^2}=-\frac{a^2}{2\lambda}$, or $\lambda=\pm\frac{ab}2$.

If $\lambda=\frac{ab}2$, then $x=-\frac aby$. Replacing $x$ with $-\frac aby$ in the third equation, you get that $y=\pm\frac b{\sqrt2}$, and then $x=\mp\frac a{\sqrt2}$. The case in which $\lambda=-\frac{ab}2$ is similar.