Max/Min Value of f(x,y) in Range x^2+y^2≤2: Find Max/Min Value of f

[MarkFL]

Good to know, Petrus!

I would highly recommend that you revisit the suggestion made by chisigma when you get time to see how his suggestion works, and how much simpler it is to use.

 

[Petrus]

Good to know, Petrus!

I would highly recommend that you revisit the suggestion made by chisigma when you get time to see how his suggestion works, and how much simpler it is to use.

I am looking at his way and some I don't understand but I need to think longer, I Will post if I got any question:)
One more question, can I Also solve this lambda way

 

[MarkFL]

Good deal! I think the more methods you can apply, the more you understand how they relate.

And yes, you have an objective function and two separate constraints, but I would get used to referring to that method as using Lagrange multipliers. (Wink)(Happy)

It just happens to traditionally use the Greek letter lambda $\lambda$.

 

[Petrus]

The problem is greatly simplified if You divide the function by 3 and the convert it in polar coordinates setting $\displaystyle x= \rho\ \cos \theta$ and $y = \rho\ \sin \theta$ so that You have to maximize/minimize the function...

$\displaystyle f(\rho, \theta)= \frac{2}{1+\rho^{2}} + \rho^{2}\ \sin \theta\ \cos \theta$ (1)

... with the condition...

$\displaystyle \frac{1}{3} \le \rho^{2} \le 2$ (2)

First we compute the partial derivatives...

$\displaystyle f_{\rho}(\rho, \theta)= \frac{4\ \rho}{(1+\rho^{2})^{2}}+ 2\ \rho\ \sin \theta\ \cos \theta$

$\displaystyle f_{\theta} (\rho,\theta)= \rho^{2}\ (\cos^{2} \theta- \sin^{2} \theta)$ (3)

... and we observe that $f_{\theta}(*,*)$ vanishes for $\displaystyle \sin \theta = \pm \cos \theta = \pm \frac{1}{\sqrt{2}}$, so that we arrive at the two equation in $\rho$...

$\displaystyle \frac{4}{(1+\rho^{2})^{2}} - 1 =0$ (4)

$\displaystyle \frac{4}{(1+\rho^{2})^{2}} + 1 =0$ (5)

The (5) has no real solutions and the only real positive solution of (4) is $\displaystyle \rho=1$, intenal os the annulus $\displaystyle \frac{1}{\sqrt{3}} \le \rho \le \sqrt{2}$ so that the point of minima or maxima are for $\displaystyle \rho=1$ and $\displaystyle \theta = \frac{\pi}{4},\ \frac{3}{4}\ \pi,\ \frac{5}{4}\ \pi,\ \frac{7}{4}\ \pi$. Further detail are left to You...

Kind regards

$\chi$ $\sigma$

So this is what I understand or don't understand:
I do understand that you turn it to polar coordinates.
I do understand that we subsitate \(\displaystyle y^2+x^2 <=>p^2\cos^2\theta+p^2\sin^2 \theta \) so we get \(\displaystyle p^2\cos^2\theta+p^2\sin^2 \theta <=> p^2(\cos^2 \theta+\sin^2\theta) <=> p^2(1)\)
when I derivate \(\displaystyle f_p\) I get same result but when I derivate \(\displaystyle f_{\theta}\) So what do I get? Here is how I do:
we can ignore that left side because it don't have any \(\displaystyle \theta\) in it so we want to focus on \(\displaystyle \rho^{2}\ \sin \theta\ \cos {\theta} \) well \(\displaystyle p^2\) is a constant so we can ignore it and only focus in \(\displaystyle \sin \theta\ \cos {\theta}\). We got a function multiplicate with a function that means we will use product rule. so I get \(\displaystyle -\cos \theta \cos\theta+sin\theta \sin\theta\) We can rewrite it nicer with \(\displaystyle \sin^2\theta-\cos^2\theta\) and we can't forget our constant \(\displaystyle p^2\) so the result is
\(\displaystyle f_{\theta} (\rho,\theta)=p^2(\sin^2\theta-\cos^2\theta)\)
and those part after I don't understand.