Find the maximum of $f(x, y) = (3x+4y)e^{\frac{-(x^2+y^2)}{2}}$ as a function of $r$ over a closed disk $x^2+y^2\leq r^2.$

Question

Find the maximum of $f(x, y) = (3x+4y)e^{\frac{-(x^2+y^2)}{2}}$ as a function of $r$ over a closed disk $x^2+y^2\leq r^2.$

So my first guess is to check the interior points and then the boundry points. But, how do I get the function $f$ as a function of $r$?

I rewrote it as following, but not sure if this is right.

$$f(x(r, t), y(r, t)) = (3r\cos{t}+4r\sin{t})e^{-\frac{r^2}{2}},$$

I can then get the parital derivatites:

$$f_r=(3\cos{t}+4\sin{t})e^{-\frac{r^2}{2}} - (3r^2\cos{t}+4r^2\sin{t})e^{-\frac{r^2}{2}}$$ $$f_t=(-3r\sin{t}+4r\cos{t})e^{-\frac{r^2}{2}}$$

Then, I can find when the derivatives are equal to $0$,

$$f_t=(-3r\sin{t}+4r\cos{t})e^{-\frac{r^2}{2}} = 0 \implies \tan{t} = \frac{4}{3}.$$

I think I'm doing this totally wrong. Maybe I should use Lagrange multipliers somehow?

Answer is: $max(r) = 5re^{\frac{-r^2}{2}}$ if $r \leq 1$ and $max(r) =\frac{5}{\sqrt{e}}$ if $r \geq 1$. $min(r) = -max(r)$.

Answer 1

You have two r's in the problem. So lets call one something different by saying $x^2+y^2=r^2\le R^2$.

You have correctly calculated $t$ which results in a maximum. This gives $\cos(t)=\frac{4}{5}$ and $\sin(t)=\frac{3}{5}$. Put this into your expression for $f_r$ and it is easy to see that $r=1$.

Now if $R>1$ then you have a maximum at $r=1$ and thus $max[f(x,y)]=5 e^{-\frac{1}{2}}$.

If $R<1$ then maximum is at $r=R$ so $max[f(x,y)]=5 R e^{-\frac{R^2}{2}}$.

You of course still need to check these are maximums.

Answer 2

First of all, let us see what happens in $D_r(0)$. We have$$f_x(x,y)=e^{-\frac{x^2+y^2}2}(-3x^2-4xy+3)\text{ and }f_y(x,y)=e^{-\frac{x^2+y^2}2}(-3xy-4y^2+4).$$Therefore, the critical points are the solutions of the system$$\left\{\begin{array}{l}-3x^2-4xy+3=0\\-3xy-4y^2+4=0.\end{array}\right.$$This system has two solutions: $\pm\left(\frac35,\frac45\right)$, which belong to $D_r(0)$ if and only if $r>1$. On the other hand,$$f\left(\pm\left(\frac35,\frac45\right)\right)=\pm\frac5{\sqrt e}.$$

Now, if $x^2+y^2=r^2$, we have $f(x,y)=(3x+4y)e^{-\frac{r^2}2}$. So, if $(x,y)=r(\cos\theta,\sin\theta)$,\begin{align}f(x,y)&=(3\cos\theta+4\sin\theta)e^{-\frac{r^2}2}\\&=5\left(\frac35\cos\theta+\frac45\sin\theta\right)e^{-\frac{r^2}2}.\end{align}So, the maximum of $f$ in that circle is $5e^{-\frac{r^2}2}$, whereas the minimum is $-5e^{-\frac{r^2}2}$.

Therefore,$$\max f=\begin{cases}\frac5{\sqrt e}&\text{ if }r\geqslant1\\\frac5{e^{\frac{r^2}2}}&\text{ otherwise }\end{cases}\text{ and }\min f=\begin{cases}-\frac5{\sqrt e}&\text{ if }r\geqslant1\\-\frac5{e^{\frac{r^2}2}}&\text{ otherwise.}\end{cases}$$

Answer 3

You got

$f(r,t)=$ $(3\cos t +4\sin t)r\exp(-r^2/2)$, $0\le t \lt 2π, 0<r\le 1.$

The problem is separable :

Find the maxima of

1)$g(t):= 3\cos t +4\sin t$, $0\le t \lt 2π$.

2) $h(r):=r\exp (-r^2/2), 0<r \le 1$.

1) $g(t)= (3,4)(\cos t, \sin t)$ (scalar product).

Note $(\cos t, \sin t)$ is a unit vector.

Hence:

$g(t)= |(3,4)||1|\cos \phi$, where $\phi$ is the angle between the vectors.

Hence

max $ (g(t))= \sqrt{3^2+4^2}=5$. $( \cos \phi =1$, i.e. $(3,4)$ and $(\cos t, \sin t)$ are parallel)

2) $h(r)=r\exp(-r^2/2)$.

$h'(r)=$ $ \exp(-r^2/2) +r(-r)\exp(-r^2/2)$;

$h'(r) >0$ for $0<r \lt 1$, i.e strictly increasing $h$.

$h'(r=1)=0$

$h'(r) < 0$ for $r > 1$ , i.e. strictly decreasing $h$.

For $r <1$ : $\max (h(r)) =r\exp(-r^2/2)$.

For $r \ge 1$: $\max (h(r))= \exp(-1/2)$.

Finally:

$\max (f(r,t)=$ $ (\max (g(t))) (\max (h(r)))$.