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.

Do you mean $x^2 + y^2 = 4$? If you actually mean the whole circle, then see [this question](https://math.stackexchange.com/questions/49473/lagrangemultiplierswithinequalityconstraints). – Robert DB May 05 '18 at 18:52

1I mean the whole circle. I am not forced to use Lagrange multiplier, so I am curious to how one should go about solving this. – StudentMaths May 05 '18 at 18:55

1Lagrange multiplier method is for $\text{equalities}$ only. Keep that in mind. – callculus42 May 05 '18 at 18:56

I guess you can use Lagrange multipliers (KarushKuhnTucker 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. – Gerhard S. May 05 '18 at 18:58
3 Answers
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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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? – StudentMaths May 05 '18 at 19:36

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.) – Robert DB May 05 '18 at 19:59
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 negativedefinite, and the hessian at $(0,2)$ is positivedefinite.
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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@StudentMaths Right you are! I've edited my answer. I hope that everything is correct now. – José Carlos Santos May 05 '18 at 19:58
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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