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

[MarkFL]

Honestly you are a really great explainer and person. It did take like 20+ post and you keep helping while you don't have to, I have always admired you for so long time. I wish their was more people like you in this world. I have had one really good math teacher that sometimes did spend his free time to learn math to other student that had problem. He did also become great friend with evryone and then their is teacher that don't bother if someone don't understand. Well it's not their problem if someone don't understand but it shows that their is people with big heart and that is something that admired. Unfortently you can't show how greatful you are in internet, but thanks a lot Mark:)
edit: I forgot to thank chisigma aswell, sorry.

Well, I truly appreciate your kind words, and I would like to say that it is a two way street. When the student demonstrates a willingness to work and a desire to learn, then the teacher is much more likely to be motivated to foster that desire in the student. Petrus, you make it easy to want to help as you clearly want to learn and you are willing to show your work and stick with it to the end. So this is really more a good reflection on you than me. I truly wish more people posting on forums were like you. (Yes)

 

[MarkFL]

So, let's gather together everything you have found so far:

You have the critical points on the interior of the annulus:

$$\left(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}} \right),\,\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}} \right)$$

and the critical points on the boundaries are?

Hint: you should find six points on each boundary.

 

[Petrus]

So, let's gather together everything you have found so far:

You have the critical points on the interior of the annulus:

$$\left(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}} \right),\,\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}} \right)$$

and the critical points on the boundaries are?

Hint: you should find six points on each boundary.

$$(1,1)$$, $$(1,-1)$$, $$(-1,1)$$, $$(-1,-1)$$, $$(\sqrt{2},0)$$, $$(-\sqrt{2},0)$$, $$(\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}})$$, $$(\frac{1}{\sqrt{6}},-\frac{1}{\sqrt{6}})$$,$$(-\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}}), $$$$(-\frac{1}{\sqrt{6}},-\frac{1}{\sqrt{6}})$$, $$(\frac{1}{\sqrt{3}},0)$$, $$(-\frac{1}{\sqrt{3}},0)$$

 

[MarkFL]

Yes, looks good. So you have 14 points at which to evaluate the function. What do you find are the absolute extrema?

 

[Petrus]

Hello,
I got now right answer did typo wrong when I did put it in calculator. max is 5 and min is -1. Thanks again for the help!:)

 

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