Find Global max/min of $f(x,y) = 2xy-x-y$ in $D$.

maxima-minimamultivariable-calculus

Find Global max/min of $f(x,y) = 2xy-x-y$ in $D$.
$D=\{(x,y)| x^2+y^2\le1, y\ge-x \}$

What I did:
I drew $D$ first, and now my plan is to find inner critical points, then find max/min points on each of part of the circle and the line, then find the value of the intersections of the circle and the line. And according to all this I decide global max/min.
(According to weierstrass theorem, $D$ is bounded and closed, which means $f$ has max and min in $D$).

  1. Critical inner points:

    $\nabla f=(2y-1,2x-1)=0 \Longrightarrow (\frac{1}{2},\frac{1}{2})$.

  2. On the line: $y=-x$
    $f(x,-x)=-2x^2 \Longrightarrow f'=-4x=0 \Longrightarrow x=0,y=0 \Longrightarrow(0,0)$.

    On the circle (And here comes the problems:
    First I tried to do with lagrange multipliers with $\nabla L = (2y-1-2x\lambda, 2x-1-2y\lambda)=0$ and $x^2+y^2=1$.
    But things got Really complicated and I couldn't find the points.

    So I decided to parametrize the circle $\vec r(t)=(\cos(t),\sin(t))$ and $f(r(t))=\sin(2t)-\cos(t)-\sin(t)$
    $f'=2\cos(2t)+\sin(t)-\cos(t)=0$ But I also got stuck here calculating this..

    Last thing I tried is $y=\pm \sqrt{1-x^2}$, and you know what happened too.

    I would really appreciate any help with how to solve these stuff.
    Thanks in advance!

Best Answer

Alternatively, convert it into polar coordinates.

$f(r, \theta) = r^2 \sin 2\theta - r \sin\theta - r\cos\theta$
$(0 \leq r \leq 1, - \frac{\pi}{4} \leq \theta \leq \frac{3\pi}{4})$

First for maximum,

We observe that at $r = 0, f(r,\theta) = 0$. So we try to find a point in the given interval where function is positive.

We have $r^2 \leq r$ and $\sin \theta + \cos\theta = \sqrt2 \sin(\frac{\pi}{4} + \theta)$ which is positive for $- \frac{\pi}{4} \leq \theta \leq \frac{3\pi}{4}$. So we must have $\sin (2\theta) \gt \sqrt2 \sin(\frac{\pi}{4} + \theta)$.

For $\theta$ in first quadrant, $\sin(\frac{\pi}{4} + \theta) \geq \frac{1}{\sqrt2}$. So we do not have a solution. For $\theta \in (-\frac{\pi}{4},0] \cup [\frac{\pi}{2},\frac{3\pi}{4}), \sin 2\theta$ is negative and so we again do not have a solution.

That leads to maximum of $0$.

Now for minimum,

$f(r, \theta) = r^2 \sin 2\theta - r \sin\theta - r\cos\theta$

$ = r^2 \cos(\frac{\pi}{2}-2\theta) - r \sqrt2 \cos(\frac{\pi}{4}-\theta)$

Writing $(\theta - \frac{\pi}{4})$ as $t$,

$ = r^2 \cos (2t) - r \sqrt2 \cos t = 2(r^2\cos^2t - \frac{r \cos t}{\sqrt2} - \frac{r^2}{2})$

$ = 2 [(r\cos t - \frac{1}{2\sqrt2})^2 - \frac{1}{8} - \frac{r^2}{2}]$ ...$(i)$

So the minimum occurs when $r$ is max (which is $1$) and $\cos (\theta - \frac{\pi}{4}) = \frac{1}{2\sqrt2}$

That leads to minimum of $ - \frac{5}{4}$.

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