Lagrange Multipliers, maximize $f=xy$ restricted to $g=x^2+y^2=r^2$

Question

So I have to solve the system of equations

$$\cases{\nabla f = \lambda \nabla g\\x^2+y^2 = r^2}.$$

Then $y=2\lambda x, x=2\lambda y$. Sorry if this is obvious, but how can I get $x$ and $y$ only as a function of $\lambda$? Otherwise I'm not able to find the values $\lambda$ to evaluate them and find a maximum.

Answer 1

You could also let $x = r \cos t, y= r \sin t$, then you have the unconstrained optimisation $\max r^2 \cos t \sin t = \max r^2 {1 \over 2} \sin (2 t) $, which is maximised at $t= { \pi \over 4}$ with value ${r ^2 \over 2}$.

Answer 2

Since $y = 2\lambda x$ and $x = 2\lambda y$, $x^2 + y^2 = 4\lambda^2(x^2 + y^2)$, i.e., $r^2 = 4\lambda^2 r^2$. Thus $4\lambda^2 = 1$, yielding $\lambda = \pm 1/2$. If $\lambda = 1/2$, then $y = x$, so from the condition $x^2 + y^2 = r^2$, we obtain $x = y = \pm r/\sqrt{2}$. If $\lambda = -1/2$, then $y = -x$, so the condition $x^2 + y^2 = r^2$ results in $x = r/\sqrt{2}$ and $y = -r/\sqrt{2}$ or $x = -r/\sqrt{2}$ and $y = r/\sqrt{2}$. The point $ (r/\sqrt{2}, r/\sqrt{2})$ (and $(-r/\sqrt{2}, -r/\sqrt{2})$) maximizes $xy$, with maximum $r^2/2$.

Answer 3

One way to proceed is to multiply your first equation by x and your second equation by y, to get

$xy=\lambda(2x^2)=\lambda(2y^2)$, $\;\;$so $2\lambda(x^2-y^2)=0$ and therefore

$y^2=x^2$ and $y=\pm x\;\;$ (since $\lambda=0\implies x=y=0$).

(Now substitute into the constraint.)