Find extrema of $f(x,y)=\sqrt{2-2x^2-y^2}$, when $2x^2-4x+y^2=-1$

Question

The task is to find the min and max. I have difficulty solving this $f(x,y)=\sqrt{2-2x^2-y^2}$, when $2x^2-4x+y^2=-1$.

Firstly, I found the domain which is $2x^2+y^2\le 2$. Then, I started with Lagrange multipliers: $$\Rightarrow\quad F(x,y, \lambda)= \sqrt{2-2x^2-y^2}-\lambda(2x^2-4x+y^2+1).$$ Made the first derivatives: \begin{align} F'_x&=-2x\bigm/\!\sqrt{2-2x^2-y^2} - \lambda(4x-4)=0 \\[0.5ex] F'_y&=-y\bigm/\!\sqrt{2-2x^2-y^2} - \lambda(2y)=0 \\[0.5ex] F'_\lambda&=2x^2-4x+y^2+1=0 \end{align} I would be very thankful if somebody help me.

Answer 1

The points at which $f$ attains its extreme points are exactly the points at which $f^2$ attains its extreme points and it easier to deal with$$F(x,y)=f^2(x,y)=2-2x^2-y^2.$$So, let $g(x,y)=2x^2-4x+y^2$ and solve the system$$\left\{\begin{array}{l}\frac{\partial F}{\partial x}(x,y)=\lambda\frac{\partial g}{\partial x}(x,y)\\\frac{\partial F}{\partial y}(x,y)=\lambda\frac{\partial g}{\partial y}(x,y)\\g(x,y)=-1.\end{array}\right.$$You will get two solutions: $\left(1\pm\frac1{\sqrt2},0\right)$, of which only $\left(1-\frac1{\sqrt2},0\right)$ belongs to the domain of $f$. So, $f$ attains its maximum there. And, of course, the minimum is $0$, which is attained at every point of the boundary of the domain.

Answer 2

There is already an accepted solution, so please consider this as a comment.

Note that we have an extremal problem for a running point $(x,y)$ constrained to live on the ellipse $$ 2(x-1)^2+y^2=1\ , $$ with axes parallel to $Ox$, $Oy$, so that $x$ runs in the interval delimited by $1\pm \frac 1{\sqrt 2}$, and respectively $y$ in the interval delimited by $\pm 1$.

On this boundary we have to (see if the following expression is defined, then) minimize / maximize $$ \sqrt{2-(2x^2+y^2)} =\sqrt{2-(-1+4x)} = \sqrt{3-4x}\ , $$ which can be easily minimized/maximized (where it makes sense).