Find the maximum of $$x_2 + y_1 - x_1 - y_2$$if $(x_1 - 1/2)^2 + (y_1 - 1/2)^2 = 1/4$ and $(x_2 - 1/2)^2 + (y_2 - 1/2)^2 = 1/4$ and $x_1y_2 = y_1x_2$ and all variables are positive.
Is there a way to do this with inequalities? What about Lagrange Multipliers?
Using Lagrange multipliers, that is to say minimizing $$F=x_2 + y_1 - x_1 - y_2+\lambda \left(\left(x_1-\frac{1}{2}\right)^2+\left(y_1-\frac{1}{2}\right)^2- \frac{1}{4}\right)+$$ $$\mu\left(\left(x_2-\frac{1}{2}\right)^2+\left(y_2-\frac{1}{2}\right)^2- \frac{1}{4}\right)+\nu(x_1y_2- y_1x_2)$$ is a pure nightmare.
What I did was to extract as much variables as I could do from the constraints that is to say $$y_{1\pm}=\frac 12\pm\sqrt{\frac 14-\left(x_1-\frac{1}{2}\right)^2}$$ $$y_{2\pm}=\frac 12\pm\sqrt{\frac 14-\left(x_2-\frac{1}{2}\right)^2}$$ $$x_2=\frac {y_2} {y_1} x_1$$ and considered all possible cases. For any combination of the $\pm$'s, $F$ is now a function of $x_1$ only and can be minimized (not pleasant but doable). Now, for each solution $x_1$, compute the value of $F$.
I shall not report all the results (too long to be typed) but, if I am not mistaken, the best point corresponds to $$x_1=\frac 14 \qquad x_2=\frac{2+\sqrt{3}}{4}\qquad y_1=\frac{2-\sqrt{3}}{4}\qquad y_2=\frac 14\qquad \implies \qquad F_{max}=\frac12$$
I let you the pleasure of doing it.