# 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

• Small detail : $-3+16+25=38$ and not $36$ – Claude Leibovici Sep 16 '18 at 4:20

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 $x=7/5$.
With a bit of arrangements I did on paper, you could show that $$f'(x) = -\dfrac{6\sqrt{-x^2+4x-3}+8x-16}{\sqrt{-x^2+4x-3}}$$ which is zero when $$6\sqrt{-x^2+4x-3}= -8x+16$$ square both sides $$-36x^2+ 144x - 108 = 64x^2 -256x + 256$$ which is $$100x^2 -400x +364 = 0$$ Roots are $$x_{1,2} = \frac{400 \pm 120}{200}$$ i.e. \begin{align} x_1 &= 2.6 \\ x_2 &= 1.4 \end{align} Either do a double derivative (which requires much time) to show that $f(x_2)$ is a maximum, or wiggle around $x_2$ if you're in a hurry.
$\bigg(\sqrt{-3+4x-x^2}+4\bigg)^2+\bigg(x-5\bigg)^2\;\forall\;x\in[1\;,3]$
Denote $x=t+2$, then: $$f(t)=\bigg(\sqrt{1-t^2}+4\bigg)^2+\bigg(t-3\bigg)^2\;\forall\;t\in[-1\;,1]\\ f'(t)=2\bigg(\sqrt{1-t^2}+4\bigg)\cdot \frac{-t}{\sqrt{1-t^2}}+2(t-3)=0 \Rightarrow\\ -2t-\frac{8t}{\sqrt{1-t^2}}+2t-6=0 \Rightarrow \\ 4t=-3\sqrt{1-t^2} \stackrel{t<0}{\Rightarrow} \\ 16t^2=9-9t^2 \Rightarrow \\ t=-\frac35.$$ Hence, for $f(t)$: $$f(-1)=32\\ f(1)=20\\ f(-\frac35)=46.08 \ \text{(max)}.$$ It means $x=t+2=-\frac35+2=\frac75$.