Show that the critical point of the function is a local minimum.

Question

Consider the function $F:\mathbb{R}^2\to\mathbb{R},(x,y)\mapsto F(x,y)=x^2+2y^2+4.97$.

$H$, hessian matrix of above function is symmetric. If $H$ is positive definite in a critical point then it is a local minimum.

Show that the critical point of item function is a local minimum.

Answer 1

$F$ has only one critical point: $(0,0)$.

For all $(x,y)$ we have for the Hessian: $H(x,y)= diag(2,4)$, which is positive definite.

Without $H$: we have $F(0,0)=0 \le F(x,y)$ for all $(x,y).$