2

As the title says I need to find extreme values(maximum and minimum) of

$$f(x,y) = xy \ln(x^2+y^2), x>0, y>0$$

I don't understand how to find critical points of this problem. I start with finding partial derivative and set derivatives equal to zero. And that is where I am stuck currently. So any help would be appreciated.

So:

$ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} (xy* ln(x^2+y^2)) = y*ln(x^2+y^2) + \frac{2xy^2}{x^2+y^2} $ $ \frac{\partial f}{\partial y} = \frac{\partial}{\partial y} (xy* ln(x^2+y^2)) = x*ln(x^2+y^2) + \frac{2x^2y}{x^2+y^2} $

So we have now:

$ \nabla f(x,y) = (0,0) $

$ y*ln(x^2+y^2) + \frac{2xy^2}{(x^2+y^2)^2} = 0 $

And

$ x*ln(x^2+y^2) + \frac{2x^2y}{(x^2+y^2)^2} = 0 $

After trying to solve these equations I get that $x=y$ Is that correct?. So I don't understand what are then critical points as It can't be (0,0)?

Adnan Selimovic
  • 133
  • 1
  • 12
  • "extrema" are maximum and minimum? – Exodd Jun 15 '17 at 18:20
  • @Exodd yes I need to find maximum and minimum. – Adnan Selimovic Jun 15 '17 at 18:27
  • Can you edit the question to include the values you found for the "first derivatives"? – Michael Seifert Jun 15 '17 at 18:41
  • Also, if you're familiar with the technique, I suspect you'll have an easier time of solving this problem if you use polar coordinates to solve this problem and then convert it back into Cartesian coordinates. – Michael Seifert Jun 15 '17 at 18:45
  • Find the maxima and the minima by solving $\nabla f=0$, then look for extreme values on the axis $x=0$, $y=0$. Recall that the points in which the function is locally extreme satisfy the equation $\nabla f=0$, but there might be points on the axis that are indeed maxima or minima because the function is restricted on a quarter of the plane, but do not satisfy the equation $\nabla f=0$. Do you have any idea on how to look for those points? – user438666 Jun 15 '17 at 18:46

3 Answers3

2

It's easier to work with polar coordinates $x = r \cos \theta, y = r \sin \theta$ where (in your case) $r > 0$ and $0 < \theta < \frac{\pi}{2}$. Then we have

$$ f(x,y) = xy \ln(x^2 + y^2) = r \cos \theta r \sin \theta \ln(r^2) = r^2 \ln r \sin(2 \theta) = g(r,\theta). $$

The partial derivatives of $g$ are given by

$$ \frac{\partial g}{\partial r} = r (2 \ln r + 1) \sin(2 \theta), \\ \frac{\partial g}{\partial \theta} = 2 r^2 \ln r \cos(2 \theta). $$

Since $r > 0$ and $\sin 2\theta > 0$ in our range, the equation $\frac{\partial g}{\partial r} = 0$ implies that $2 \ln r + 1 = 0$ or $r = e^{-\frac{1}{2}}$. The second equation then implies that $\cos 2 \theta = 0$ so $\theta = \frac{\pi}{4}$. Hence, we have one critical point at $r = e^{-\frac{1}{2}}$ and $\theta = \frac{\pi}{4}$. The second partial derivatives of $g$ are given by

$$ \frac{\partial^2 g}{\partial r^2} = (2 \ln r + 3) \sin(2 \theta), \\ \frac{\partial^2 g}{\partial r \partial \theta} = 2 r(2 \ln r + 1) \cos(2 \theta), \\ \frac{\partial^2 g}{\partial \theta} = -4 r^2 \ln r \sin(2 \theta). $$

Hence, the Hessian at $r = e^{-\frac{1}{2}}, \theta = \frac{\pi}{4}$ is given by

$$ \begin{pmatrix} 2 & 0 \\ 0 & 2e^{-1} \end{pmatrix}. $$

Since the Hessian is positive definite, $(e^{-\frac{1}{2}}, \frac{\pi}{4})$ a local minimum point for $g$ (and so $\frac{e^{-\frac{1}{2}}}{\sqrt{2}} \left( 1, 1 \right)$ is a local minimum for $f$).

levap
  • 65,634
  • 5
  • 79
  • 122
  • 2
    The point is that while the formula for the gradient and the Hessian changes when we change coordinates, the critical points and the signature of the Hessian can be computed using the regular formulas. – levap Jun 15 '17 at 18:54
  • Can you explain me how you get that $ \theta < \frac{\pi}{4} $ ? – Adnan Selimovic Jun 15 '17 at 19:49
  • @AdnanSelimovic: Sorry, it should be $0 < \theta < \frac{\pi}{2}$ because $x > 0, y > 0$ means we are in the first quadrant. – levap Jun 15 '17 at 20:30
2

We look for points for wich $\nabla f=0$

The method you describe is this:

$\dfrac{\partial f}{\partial x}=y\ln(x^2+y^2)+xy\dfrac{2x}{x^2+y^2}$

$\dfrac{\partial f}{\partial y}=x\ln(x^2+y^2)+xy\dfrac{2y}{x^2+y^2}$

(You need to practice the derivatives :)

Setting the partials to zero we can get the coordinates for those points.

$y\ln(x^2+y^2)+xy\dfrac{2x}{x^2+y^2}=0$

$x\ln(x^2+y^2)+xy\dfrac{2y}{x^2+y^2}=0$

$\dfrac{y}{x}\ln(x^2+y^2)=\dfrac{x}{y}\ln(x^2+y^2)\implies x^2=y^2\;,x^2+y^2\neq1$

Then, $y=x$ or $y=-x$, dropping the second as $x\gt0$ and $y\gt0$

$x\ln(2x^2)=-\dfrac{2x^3}{2x^2}\implies \ln(2x^2)=-1$ or $x=\pm\dfrac{e^{-1/2}}{\sqrt{2}}$

Dropping the negative solution:

$x=y=\dfrac{e^{-1/2}}{\sqrt{2}}$

(We need to consider the case $x^2+y^2=1$. It's easily seen that the partials doesn't vanish: for those points, $2x^2y=0$, that is not possible with $x\gt0\;,y\gt0$)

Rafa Budría
  • 7,364
  • As of derivates I have mixed it with partial derivatives of second order because I've stuck on this problem for couple hours. Thanks for the tip anyway, I edited it now. So basically when I get, like in this problem, that x=y I just plug in one equation and I get the solution. Haha! I don't know why I have not done this. It's that simple. Thank you so much! – Adnan Selimovic Jun 15 '17 at 19:19
  • That time is not wasted time, believe me. And you got three different approaches! :) – Rafa Budría Jun 15 '17 at 19:25
1

In polar coordinates, you have $$\sin(\theta)\cos(\theta)r^2\ln(r^2)$$

$\sin(\theta)\cos(\theta)$ is maximal when $\theta=\pi/4$, and $r^2\ln(r^2)$ is unbounded above. So there is no maximum on the specified domain.

$\sin(\theta)\cos(\theta)$ is always positive on the specified domain, and $r^2\ln(r^2)$ does have a negative absolute minimum value. So by finding the minimum of $r^2\ln(r^2)$ and pairing that with $\theta=\pi/4$, you can find the absolute minimum using one-variable calculus.

With $\frac{d}{dr}\left(r^2\ln(r^2)\right)=2r\left(\ln(r^2)+1\right)$, find that the minimal $r$-value is $e^{-1/2}$.

2'5 9'2
  • 54,717