Find the absolute maximum and minimum of $f(x,y) = 1-xy$ in the domain $x^2+y^2 \le 2$

Question

My exam is tomorrow and I can't figure this out. I tried using Lagrange and it doesn't seem to work (I'm left with an unsolvable system of equations)

Answer 1

I would use polar coordinates $x=r\cos t$, $y=r\sin t$

Your problem can be rewritten as maximize/minimize $f(r,t)=1-r^2\sin t\cos t=1-\frac{1}{2}r^2\sin 2t$ with $0<r<\sqrt 2$ and $0\le t< 2\pi$

The maximum is $2$ is attained when $r=\sqrt 2$ and $2t=\frac{3\pi}{2}$ or $\frac{3\pi}{2}+2\pi$, and the minimum is $0$ and is attained when $r=\sqrt 2$ and $2t=\frac{\pi}{2}$ or $\frac{\pi}{2}+2\pi$

Answer 2

First you should calculate $\nabla f(x,y)=(-y,-x)=(0,0)$. So $(0,0)$ is the minimum, now you have to use lagrande multiplier to the constraint $x^2+y^2=2$ using $g(x,y)=x^2+y^2-2$

Answer 3

I think this is easy enough to not have to invoke Lagrange. Fix a point $(x_0,y_0)$ in $D=\{(x,y):x^2+y^2\leq2\}$ with non-negative first coordinate and observe that for $\varepsilon\geq0$ $$ 1-x_0(y_0+\varepsilon)\leq1-x_0y_0\leq1-x_0(y_0-\varepsilon). $$ Translated in terms of $f$ this tells you that both the maximum and minimum of $f$ in $D\cap\{(x,y):x\geq0\}$ are attained (particularly) in the circle $\partial D$. The same reasoning (with reversed inequalities in the above estimates) can be made for a point with negative first coordinate, allowing us to conclude that the maximum and minimum of $f$ in $D$ are attained in the circle $\partial D$. This reduces the problem to find the maximum and minimum of $$ f(\sqrt{2}\cos\theta,\sqrt2\sin\theta)=1-2\cos\theta\sin\theta,\,\,\,\,\theta\in[0,2\pi), $$ which I leave to you.