Explaining why the $F(x,y)=x^2+y^2$ has a minimum on $xy = 3$ but not a maximum

Question

I have been asked to find the minimum value of the function $F:\Bbb{R}^2\to\Bbb{R}$, $F(x,y)=x^2+y^2$ on the curve $xy = 3$.

I found the minimum value with using Lagrange multipliers theorem. The answer is $(-\sqrt{3},-\sqrt{3})$ and $(\sqrt{3}, \sqrt{3})$; and when you place these into the function, the value is $6$, so the minimum value is $6$.

There are two problems now:

1. In class the only way we learn to say if its minimum or maximum is to use weirshtrass theorem that saying that the function gets max and min and than choose between the potential max and min points. Here, i can't use weirshtrass theorem so i am not sure how i am supposed to prove this is a minimum point and not maximum point. I thought of taking another random point on the curve and just say that it's smaller than it but it's look not formal and strong.

2. I need to explain why the function has minimum on the curve but does not have a maximum on the curve. I don't know what they want from me and how I am supposed to approach it.

At the bottom line, I think they miss to teach us all we need in class, so I am asking for explanation or a link to a video with explanation to this problem.

Answer 1

For the maximum, you can prove that F isn't bounded on the curve xy=3.

For any bound $M \geq 6$, we have with $(x,y) := (M,\frac{3}{M})$ that $xy=3$ and $f(x,y) = x^2 + y^2 = M^2 + \frac{9}{M^2} > M$

Answer 2

I edit my post to try to make the argument more clear.

$1$ variable case.

Claim: take $f:\mathbb{R} \to \mathbb{R}$ continuous such that $\lim_{x\to{-\infty}}f(x)=\lim_{x\to +\infty}f(x)=+\infty$ then $f$ has a minimum (not a maximum obviusly), this property is called coercivity. An example of coercive function is $ f_1(x)=x^2$ but also $f_2(x)=x^2+\arctan(45+\sin(x^{2/5}))$. They both explode to $+ \infty $ as $|x|$ grows. You can intuitively see that the minimum cannot be found far away from $0$ because they will be grater and grater as we move away. Rigorously take $f(0)=c$ the definitions of limits give us that there exists a constant $R>0$ such that if $x>R$ then $f(x)>c$ , and if $x<-R$ $f(x)>c$. Therefore the $$\inf_\mathbb{R} f(x)=\inf_{[-R,R]}f(x)$$ the second infimum exists and is a minimum by Weierstass because we have restricted to a compact set ( we are not so far away from $0$).

2 variable case

The idea is the same, you have to observe that the function grows as you move away from the origin. For example you find that there is one point where the function attains the value $6$, now you can search the minimum only in the set $\{xy=3\} \cap \{x^2+y^2 \le 6\}$ which is compact!

In this case compactness is easy and I see now that coercivity is not needed, altought if you want to maximize a generic function $G(x,y)$ over $\mathbb{R}^2$ you need precisely this property.

Answer 3

Convert this into polar coordinates,

$x = r\cos\theta, y = r\sin\theta$

Curve $xy = 3 \implies r^2 \sin2\theta = 6$

We need to find min and max of $ \displaystyle f(r, \theta) = r^2 = \frac{6}{\sin2\theta} $

As $r \in \mathbb{R}, \sin2\theta \geq 0$ and $f$ is minimum when $\sin2\theta$ is maximum which is $1$.

$\therefore \min f(r, \theta) = 6$

Also you can clearly see there is no maximum as $\theta \to n\pi/2, f \to \infty$.