Global extrema of $f(x,y)=\frac{x^2}{2}+3y^2$ on the set $M(g):=\{(x,y)\in\mathbb{R}^2\mid g(x,y)=0\}$

Question

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

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$M(g)$ 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 $$f(0,\sqrt{5}) = 15,f(0,-\sqrt{5})=15,\\ f(5,0)= \frac{25}{2},f(-5,0)= \frac{25}{2},\\ f(4,\sqrt{3})= 17,f(-4,\sqrt{3})= 17,f(4,-\sqrt{3})= 17,f(-4,-\sqrt{3})= 17.$$ 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?

Answer 1

You are misunderstanding how a necessary condition works, I think.

In this problem, once you have found the set of ordered pairs from the method of Lagrange multipliers, you can be sure that the global maximizer and minimizer are in this set (if they exist). This is because being in this set is a necessary condition for being the global maximizer or minimizer. This condition isn't sufficient for being a global maximizer or minimizer - this means that it is not true that every ordered pair in the set corresponds to a global maximizer or minimizer.

Since we know that the global maximizer and minimizer are in this set of ordered pairs, we can simply plug all these points into the function and see what values we get. Whichever one gives the biggest (resp. smallest) value corresponds to the global maximizer (resp. minimizer).

Answer 2

The Lagrange multiplier method says that IF you are at a extreme value, it must be at a critical point. So that by itself does not guarantee they exist.

However, as you noted, the domain is compact and the function is continuous, so the extreme value theorem guarantees existence. So once you know they exist and they must be at one of the multiplier points, calculating the values there is sufficient.

Answer 3

Since you know that the global min/max exist and that they must be selected from the list of critical points of the Lagrangian, you just need to compute $f$ over each of those critical points and observe where the min/max values are attained.

Don't forget that you need to check that the rank of the Jacobian matrix of the restrictions is maximal.