Find global maximum and global minimum of
$$f(x,y) = \frac{\ln \left(\frac54y^2+(x-2)^2 \right)}{\sqrt[3]y}$$
over
$$D = \left\{ (x,y) \in \mathbb{R}^2 :\frac14 y^2 + (x-2)^2 = 1 \right\}$$
My textbook says the minimum is at $(2;-2)$ and the maximum at $(2;2)$, but I don't know how to prove it. I tried to draw level curves with no success .Then I tried with Lagrange multipliers test and it doesn't work.
Any suggestions?
On
With Lagrange multipliers.
$$ L = \frac{\ln \left(\frac54y^2+(x-2)^2 \right)}{\sqrt[3]y}+\lambda \left(\frac14 y^2 + (x-2)^2 - 1\right) $$
The stationary points are determined by the solutions to
$$ \nabla L = 0 = \left\{ \begin{array}{l} 2 (x-2) \left(\lambda \sqrt[3]{y} \left(4 (x-2)^2+5 y^2\right)+4\right) \\ 3 y^2 \left(\lambda \sqrt[3]{y} \left(4 (x-2)^2+5 y^2\right)+20\right)-2 \left(4 (x-2)^2+5 y^2\right) \ln \left((x-2)^2+\frac{5 y^2}{4}\right) \\ y^2+4 (x-4) x+12 \\ \end{array} \right. $$
Analyzing those equations we can observe that $x=2$ gives
$$ 0=\left\{ \begin{array}{l} 0 \\ 3 y^2 \left(5 \lambda y^{7/3}+20\right)-10 y^2 \log \left(\frac{5 y^2}{4}\right) \\ y^2-4 \\ \end{array} \right. $$
now with $y = 2$ we have
$$ 0 = \left\{ \begin{array}{l} 0 \\ 12 \left(20 \sqrt[3]{2} \lambda +20\right)-40 \log (5) \\ 0 \\ \end{array} \right.\Rightarrow \lambda = \frac{\log (5)-6}{6 \sqrt[3]{2}} $$
and for $y = -2$ we have
$$ 0=\left\{ \begin{array}{l} 0 \\ 12 \left(-20 \sqrt[3]{2} \lambda +20\right)-40 \log (5) \\ 0 \\ \end{array} \right.\Rightarrow \lambda = -\frac{\log (5)-6}{6 \sqrt[3]{2}} $$
thus obtaining as stationary points $(2,\pm 2)$
NOTE
To consider the solutions with the branch obtained from $ \left(\lambda \sqrt[3]{y} \left(4 (x-2)^2+5 y^2\right)+4\right)=0$ we proceed as follows: considering $z = 4 (x-2)^2+5 y^2$ we have
$$ \left\{ \begin{array}{rcl} \lambda \sqrt[3]{y} z+4 & = &0\\ 3 y^2 \left(\lambda \sqrt[3]{y} z+20\right)-2 z \log \left(\frac{z}{4}\right) & = & 0\\ 4 y^2-z+4 & = & 0\\ \end{array} \right. $$
Solving for $y,z,\lambda$ we obtain $z = 1589.53, y = \pm 19.9094$ and now solving
$$ \cases{5y^2+4(x-2)^2= 1589.53\\ y =\pm 19.9094 } $$
we obtain in any case, complex values for $x$.
Let $\bar{x} := x-2$ and $\bar{y} := 0.5y$. Then, the objective function and feasible region of your transformed problem are $f(\bar{x},\bar{y}) := \frac{\text{ln}(5\bar{y}^2 + \bar{x}^2)}{\sqrt[3]{2\bar{y}}}$ and $\bar{D} := \left\{ (\bar{x},\bar{y}) \: : \: \bar{x}^2 + \bar{y}^2 = 1\right\}$.
Next, use the transformation $\bar{y} := \sin(\theta)$ and $\bar{x} := \cos(\theta)$ for $\theta \in [0,2\pi]$. The objective function is transformed to $f(\theta) = \frac{\text{ln}(1 + 4\sin^2(\theta))}{\sqrt[3]{2\sin(\theta)}}$, and the constraint is $\theta \in [0,2\pi]$.
Finally, use the transformation $z := \sqrt[3]{\sin(\theta)}$. The constraint now is $z \in [-1,1]$ and the objective function is $f(z) = \frac{1}{\sqrt[3]{2}} \frac{\text{ln}(1 + 4z^6)}{z}$. I will let you prove that $f$ is a monotonically increasing function of $z$ on this interval, which implies that the minimum is attained at $z = -1$ and the maximum is attained at $z = +1$.
Backtracking, we get that the minimum is attained at $(x,y) = (2,-2)$ and the maximum is attained at $(x,y) = (2,2)$.