Multivariable calculus, extreme values

Question

I have been trying to solve this problem:

For $(x,y) \in \mathbb{R}^2$, find the minimum and maximum value of $f\left( x,y\right) =x\sqrt{6}-3y$ on $S=\left\{ \left( x,y\right) \in \mathbb{R} ^{3}:3\leq x\leq 6,x^{2}-6x-2y^{2}+8y\geq 0\right\}$

Alternatively, $S$ can be written as $$S=\left\{ \left( x,y\right) :3\leq x\leq 6;\left( x-3\right) ^{2}-\dfrac{\left( y-2\right) ^{2}}{\dfrac{1}{2}}\geq 1\right\} $$ and represented as the intersection between the blue and red areas

We can see that $\nabla f(x,y)\neq 0\ \forall (x,y) \in \mathbb{R}^2$, so the extremes are at the edge of the domain. Here is difficult to solve for $y$ and plug it and differentiate $f(y)|_{\partial S}$. I'm hoping to find a way to solve it without using Lagrange multipliers.

Answer 1

For each $(x,y)\in\Bbb R^2$ with $x\geqslant3$, you have$$x^2-6x-2y^2+8y=0\iff x=3+\sqrt{2y^2-8y+9}.$$So, you have two cases to consider:

• $x=6$: then $0\leqslant y\leqslant4$ and $f(x,y)=6\sqrt6-3y$, whose maximal value ($6\sqrt6$) is attained when $(x,y)=(6,0)$ and whose minimal value ($6\sqrt6-12$) is attained when $(x,y)=(6,4)$.
• $x=3+\sqrt{2y^2-8y+9}$: then, again, $0\leqslant y\leqslant4$ and$$f(x,y)=\sqrt{6} \left(\sqrt{2 y^2-8 y+9}+3\right)-3 y.\tag1$$So, let $\varphi(y)$ be the RHS of $(1)$. Then$$\varphi'(y)=\frac{2 \sqrt{6} (y-2)}{\sqrt{2 y^2-8 y+9}}-3\quad\text{and}\quad\varphi'(y)=0\iff y=2+\frac{\sqrt6}2.$$Furthermore, $\varphi'(y)<0$ is $y<2+\frac{\sqrt6}2$ and $\varphi'(y)>0$ if $y>2+\frac{\sqrt6}2$. Therefore, the minimum of $\varphi$ is$$\varphi\left(2+\frac{\sqrt6}2\right)=-6+7\frac{\sqrt6}2<6\sqrt6-12.$$

So, the minimum of $f$ in that region is $6\sqrt6$ (attained at $(6,0)$) and the minimum is $-6+7\frac{\sqrt6}6$ (attained at $\left(5,2+\frac{\sqrt6}2\right)$).