extrema(max-min) two variable constrained function $f(x,y)=x^2+y^2$ with "fixed" Hessian matrix

Question

I have this function $f(x,y)=x^2+y^2$ under the constrain : $x^6+3y^2=1$

I use the Lagrange multiplier method :

$$ \left\{ \begin{array}{} 2x=6x^5\lambda\\ 2y=6y\lambda\\ x^6+3y^2-1=0 \end{array} \right. $$

We can easily get the points : $(\pm 1,0),(0,\pm \frac{1}{\sqrt{3}})$

Question : How do I evaluate if a point is max or min ? if the hessian matrix is "fixed" like so :

$$H= \begin{bmatrix} 2 & 0 \\ 0 & 2 \\ \end{bmatrix} $$

The determinant is always positive. We can even see both positive eigenvalues. But that does not mean that every point is a minimum! (right ?)

The only thing I can possibly think is to plug the points into the function, and check how is greater or not.

$f(\pm 1,0)=1$

$f(0,\pm \frac{1}{\sqrt{3}})=\frac{1}{3}$

So Just with intuition $1$ si greater than $\frac{1}{3}$ so $(\pm 1,0)$ are points of maximum; and so $(0,\pm \frac{1}{\sqrt{3}})$ are points of minimum.

But How do I know this is the right reasoning? Like, if I had found $4$ critical points how would I classified them? Are there other methods which I can apply?

Answer 1

Since the set $\{(x,y)\in\mathbb{R}^2\,|\,x^6+3x^2=1\}$ is compact, $f$ must have a maximum and a minimum there. And it must be attained at some of the points that you got. So:

• the maximum is $1$ and it is attained at $(\pm1,0)$;
• the minimum is $\frac13$ and it is attained at $\left(0,\pm\frac1{\sqrt3}\right)$.