Find the absolute max and min of a multivariable function on a bounded by a circle?

Question

So i do understand everything up the square rectangle, in the photo here i mean, how did he come up with $(±2,0), (0,±1)$ is it because of $g(2cos x, sin x)$ and if that is the case why would he compute the function after parametrizing it DOES NOT MAKE SENSE. please help me understanding the answer as i am having an exam on it today in two hours.

Answer 1

After parameterization of the ellipse by $(2\cos t, \sin t)$, $t\in [0,2\pi)$, the target function depends only on the parameter $t$, i.e., $$ g(2\cos t, \sin t) = (3\cos ^2t+1)e^{-4} $$ so, in order to find critical points you should take the derivative w.r.t. $t$, $$ g'(t) = -6e^{-4}\cos t \sin t = 0, $$ clearly $t\in \{0,\pi/2, \pi, 3\pi/2\}$ in order to satisfy the equation. Plugging these $t_0$'s into $(2\cos t, \sin t)$ will yield $(\pm2, 0), (0, \pm 1).$

Another way that you can use is just to plug in $g(x,y)$ the ellipse formula and get $g(y) = (4-3y^2)e^{-4}$, and then proceed likewise and get the points $(\pm2,0)$.

Third way, which is no recommended in this case, is to use Lagrange multipliers.