maximum value of expression $(\sqrt{-3+4x-x^2}+4)^2+(x-5)^2$ maximum value of $\bigg(\sqrt{-3+4x-x^2}+4\bigg)^2+\bigg(x-5\bigg)^2\;\forall\;x\in[1\;,3]$
what i try
$\displaystyle -3+4x-x^2+16+8\sqrt{-3+4x-x^2}+x^2+25-10x$
$\displaystyle -6x+38+8\sqrt{-3+4x-x^2}$
using derivative it is very lengthy
help me how to solve, thanks in advance

Let $x = 2+\cos \theta$ where $0\leq \theta \leq \pi$. Then the equation is equivalent to 
$$
(\sin\theta + 4)^{2} + (\cos\theta-3)^{2} = 26 + 8\sin\theta - 6\cos\theta = 26 + 10\sin(\theta+\alpha)
$$
where $-\pi/2 < \alpha <0$ satisfies $\tan \alpha = -3/4$. This attains minimum at $\theta = \pi/2 - \alpha$, which corresponds to $x=7/5$.  
