# Show that $f$ gets a global minimum and a global maximum on $D$

Let $$D=\{(x,y)\in \mathbb{R}^2\mid x^2+y^2\leq 1\}$$ and $$f: D \rightarrow \mathbb{R}$$ with $$f(x,y)=3x^2-2xy+3y^2$$.

(a) Show that $$f$$ gets a global minimum and a global maximum on $$D$$.

(b) Determine all points on $$D$$ where $$f$$ gets its global minimum and global maximum.



For question (b) first I found the critical points on the boundary of the circle $$x^2+y^2=1$$ using Lagrange multipliers. Then I determined the critical points of $$f(x,y)$$ for the interior of the circle. Then combining all results I foundthe global minimum and the global maximum.

But how can we show (a) ? Could you give me a hint?

• The complement is $x^2+y^2 >1$. Commented Jun 19, 2021 at 17:00
• @MaryStar If $g$ is a continuous function and $U$ is an open set, then $g^{-1}(U)$ is open. (The same applies with the word open replaced in both places by closed). Here the set $U=(1,\infty)$ is open and $g(x,y) = x^2+y^2$. Commented Jun 19, 2021 at 17:05
• @MaryStar I did not say that, it is clearly not equal. It is $g^{-1}((1,\infty))$. Commented Jun 19, 2021 at 17:10
• @MaryStar Yes. You could also note that $g^{-1}([0,1]) = D$ which is a little more direct. Since $g$ is continuous and $[0,1]$ closed then so is $D$. Commented Jun 19, 2021 at 17:14