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I am asked to find the minimum and maximum value to the function $f(x,y) = \frac{y}{x^2+y^2+4}$ on the the circle $x^2 + y^2 ≤ 4$. When I try to use Lagrange multiplier I get some very nasty equations.

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  • $\begingroup$ Do you mean $x^2 + y^2 = 4$? If you actually mean the whole circle, then see this question. $\endgroup$ – rwbogl May 5 '18 at 18:52
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    $\begingroup$ I mean the whole circle. I am not forced to use Lagrange multiplier, so I am curious to how one should go about solving this. $\endgroup$ – StudentMaths May 5 '18 at 18:55
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    $\begingroup$ Lagrange multiplier method is for $\text{equalities}$ only. Keep that in mind. $\endgroup$ – callculus May 5 '18 at 18:56
  • $\begingroup$ I guess you can use Lagrange multipliers (Karush-Kuhn-Tucker theorem). But you could alternatively first look for boundary extrema and then for interior extrema. On the boundary you have $x^2+y^2=4$, that is, you need to maximize/minimize $y/8$ on the circle. In the interior you don't need a Lagrange multiplier. $\endgroup$ – Gerhard S. May 5 '18 at 18:58
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You know that $f$ is continuous everywhere, so it will have extrema on the (compact) disk. You can check the interior for critical points $(\nabla f = 0)$, and then (if necessary) the boundary with Lagrange multipliers.

The equation $\nabla f = 0$ has solutions $(x, y) = (0, \pm 2)$. Extrema have to occur at critical points. Where are these points located in relation to the circle?

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  • $\begingroup$ These points satisfies x^2+y^2 ≤ 4 which means that they are on the boundary of the disk? So there are no max/min on the interior? $\endgroup$ – StudentMaths May 5 '18 at 19:36
  • $\begingroup$ Right, so Lagrange multipliers are unnecessary. The function takes on its extrema on the circle already. (Or rather, that's where the extrema could happen. The other answers show that this is in fact where they happen.) $\endgroup$ – rwbogl May 5 '18 at 19:59
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To begin with, you determine the points at wich the gradient of $f$ is $0$; these are $(0,\pm2)$. The hessian at $(0,2)$ is negative-definite, and the hessian at $(0,-2)$ is positive-definite.

At the boundary of the disk, $f(x,y)=\frac y8$. So, the minimum there is $-\frac12$ and the maximum is $\frac12$.

Since $f(0,2)=\frac14$ and $f(0,-2)=-\frac14$, the maximum of $f$ is $\frac12$ and the minimum is $-\frac12$.

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  • $\begingroup$ Thank you very much for this! Should it be f(x,y) = y/8? $\endgroup$ – StudentMaths May 5 '18 at 19:34
  • $\begingroup$ @StudentMaths Right you are! I've edited my answer. I hope that everything is correct now. $\endgroup$ – José Carlos Santos May 5 '18 at 19:58
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Change variables to

$$ x = \rho \cos\theta\\ y = \rho \sin \theta $$

and then

$$ L(\rho,\theta,\epsilon) = \frac{\rho\sin\theta}{\rho^2+4}+\lambda( 2-\rho-\epsilon^2) $$

Here $\epsilon$ is a slach variable to avoid the inequality.

The solutions are

$$ \rho = 0 \rightarrow x = y = 0\\ \rho = 2, \theta = \pm \pi/2 $$

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