Is it possible to use Lagrange multiplier to solve this problem?

Question

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.

Answer 1

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?

Answer 2

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 $-\frac12$.

Answer 3

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

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