Problem
Given $2$ points $A(0,3)$ and $B(4,0)$ on the plane, and a point $Q$ on the circle $x^2+y^2=4$.
Find the minimum value of $\frac{2}{3}\overline{QA} + \overline{QB}$.
My Idea
Let $Q$ be $(2\cos{t},2\sin{t})$. Then \begin{align*} \frac{2}{3}\overline{QA} + \overline{QB} &= \frac{2}{3}(\sqrt{(2\cos{t}-0)^2+(2\sin{t}-3)^2})+\sqrt{(2\cos{t}-4)^2+(2\sin{t}-0)^2} \\ &= \frac{2}{3} \sqrt{13-12 \sin{t}} + 2 \sqrt{5-4 \cos{t}} \end{align*} Taking the derivative to find the mimimum: $$\frac{-4 \cos{t}}{\sqrt{13 - 12 \sin{t}}} + \frac{4 \sin{t}}{\sqrt{5 - 4 \cos{t}}} = 0$$ However the computation is complicated. Are there another ways to deal with this problem?
You were on the right track. It is geometrically pretty evident that the minimizing $Q$ lies in the first quadrant, so you just have to solve $$ \frac{\sin t}{\sqrt{5-4\cos t}}=\frac{\cos t}{\sqrt{13-12\sin t}}\tag{1}$$ for $t\in\left(0,\frac{\pi}{2}\right)$. With such constraint $(1)$ boils down to $$ \sin^2(t)(13-12\sin t) = \cos^2(t)(5-4\cos t)\tag{2} $$ which can be factored as $$ (3\sin(t)+\cos(t)-2)(3\sin(t)-\cos(t)+2\cos(2t))=0.\tag{3}$$ The term $3\sin(t)-\cos(t)+2\cos(2t)$ is strictly positive for $t\in\left(0,\frac{\pi}{2}\right)$, hence the solution is provided by the solution of $3\sin(t)+\cos(t)=2$, i.e. by $$ t=\arccos\left(\frac{2+3\sqrt{6}}{10}\right)\approx 20^\circ 47'48''.\tag{4}$$ The minimum value of $\frac{2}{3}QA+QB$ turns out to be $\frac{4}{3}\sqrt{10}$.