Finding critical points in a set

Question

Consider the function $f(x, y) = xy$ on the set $S = \{(x,y)\in R^2 | x^2 + 4y^2 ≤ 1\}$.

Find the critical points of f in the interior of the set $S$.

I understand that to find the critical points you simply make the gradient equal to 0 but I don't understand how to find the ones in the interior of the set. Please help. Thanks.

Answer 1

This problem can be analyzed from two steps:

1. Judge whether the critical points are in the interior or not. Then we need to solve the problem: $$\min/max f(x,y)=xy\\\text{s.t.} x^2+4y^2<1$$ In terms of our purpose, just calculate the gradient and let it equal to 0, i.e., $\nabla f=(y,x)=0$. Then (0,0) is in the interior(if not, we can conclude that the critical points are not in the interior) and may be a critical point. Now we need to calculate $$A=\frac{\partial^2 f}{\partial x^2}(0,0)=0\\B=\frac{\partial^2 f}{\partial x\partial y}(0,0)=1\\C=\frac{\partial^2 f}{\partial x^2}(0,0)=0$$ Because $AC-B^2=-1<0$, (0,0) is not the critical point.
2. Judge whether the critical points are in the boundary.Now the problem is converted to：$$\min/\max f(x,y)=xy\\\text{s.t.} x^2+4y^2=1$$ Using Lagrange multipler method, we need to find the critical points of the equation:$$g(x,y)=f(x,y)-\lambda (x^2+4y^2-1)$$ Make its gradient equal to 0: \begin{equation} \begin{cases} \frac{\partial g}{\partial x}&=y-2\lambda x=0\\ \frac{\partial g}{\partial y}&=x-8\lambda y=0\\ \frac{\partial g}{\partial\lambda}&=1-x^2-4y^2=0\\ \end{cases} \end{equation} It's easy to get the result that: 1) $\lambda=\frac{1}{4}$,$x=\frac{1}{\sqrt{2}},y=\frac{1}{2\sqrt{2}}$ or $x=-\frac{1}{\sqrt{2}},y=-\frac{1}{2\sqrt{2}}$; 2) $\lambda=-\frac{1}{4}$,$x=-\frac{1}{\sqrt{2}},y=\frac{1}{2\sqrt{2}}$ or $x=\frac{1}{\sqrt{2}},y=-\frac{1}{2\sqrt{2}}$.

Based on 1 and 2, we can conclude that the critical points are in the boundary and they are $(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}),(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})$ and $(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}),(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})$.