Find global extrema of $f(x,y)=\frac{{x}^{2}}{2}+3{y}^{2}$ on the set $M(g):=\{(x,y)\in {\mathbb{R}}^{2}\mid g(x,y)=0\}$ where $g(x,y)={x}^{2}+{y}^{4}-25$.

$M(g)$ is compact and $f$ continuous so we know that there must exists a global maximum and global minimum. The conditions of the Lagrange multiplier methods are satisfied and if we solve the equations that result from the Lagrange multiplier method we get the following points:

$(0,\sqrt{5}),(0,-\sqrt{5}),(5,0),(-5,0),(4,\sqrt{3}),(-4,\sqrt{3}),(4,-\sqrt{3}),(-4,-\sqrt{3}).$

and

$f(0,\sqrt{5})=15,f(0,-\sqrt{5})=15,\phantom{\rule{0ex}{0ex}}f(5,0)=\frac{25}{2},f(-5,0)=\frac{25}{2},\phantom{\rule{0ex}{0ex}}f(4,\sqrt{3})=17,f(-4,\sqrt{3})=17,f(4,-\sqrt{3})=17,f(-4,-\sqrt{3})=17.$

So far so good.

However our sample solution says that "from the above values we see that $\frac{25}{2}$ is the global minimum and 17 the global maximum. "

As far as I have understood the Lagrange multiplier method it only delivers a necessary but not sufficient condition. So we don't know if one of the three points $\frac{25}{2},15,17$ is a saddle point. To make sure that the points are indeed extrema we have to resort to another method (e.g. plug in the condition into $f$).

Am I am right or is there something I don't see or didn't understand correctly?

$M(g)$ is compact and $f$ continuous so we know that there must exists a global maximum and global minimum. The conditions of the Lagrange multiplier methods are satisfied and if we solve the equations that result from the Lagrange multiplier method we get the following points:

$(0,\sqrt{5}),(0,-\sqrt{5}),(5,0),(-5,0),(4,\sqrt{3}),(-4,\sqrt{3}),(4,-\sqrt{3}),(-4,-\sqrt{3}).$

and

$f(0,\sqrt{5})=15,f(0,-\sqrt{5})=15,\phantom{\rule{0ex}{0ex}}f(5,0)=\frac{25}{2},f(-5,0)=\frac{25}{2},\phantom{\rule{0ex}{0ex}}f(4,\sqrt{3})=17,f(-4,\sqrt{3})=17,f(4,-\sqrt{3})=17,f(-4,-\sqrt{3})=17.$

So far so good.

However our sample solution says that "from the above values we see that $\frac{25}{2}$ is the global minimum and 17 the global maximum. "

As far as I have understood the Lagrange multiplier method it only delivers a necessary but not sufficient condition. So we don't know if one of the three points $\frac{25}{2},15,17$ is a saddle point. To make sure that the points are indeed extrema we have to resort to another method (e.g. plug in the condition into $f$).

Am I am right or is there something I don't see or didn't understand correctly?