Maximize this

Let $y=\sqrt{x}+4\sqrt{1-\frac{x}{2}} = \sqrt{x}+\sqrt{16-8x}$ Differentiating, we get

$\frac{dy}{dx}=\frac{1}{2\sqrt{x}}-\frac{4}{\sqrt{16-8x}}=0$

Solving, we get $x=\frac{2}{9}$ . Substituting in, we get the answer.

The condition of x is $x\in[0;2]$ Called the expression is A, we see that $A=\sqrt{x}+4\sqrt{1-\frac{x}{2}}=\sqrt{x}+\sqrt{2^3}.\sqrt{2-x}$ Using Cauchy-Schwarz inequality, we have $A^2\leq(1+8)(x+2-x)$ $\Leftrightarrow A^2\leq 18$ $\Leftrightarrow A\leq3\sqrt{2}\approx 4.24$ The equality holds when $x=\frac{2}{9}$

Using Cauchy-Schwarz inequality as follows:

$\begin{aligned} \left(\sqrt x + 2 \sqrt{1-\frac x2} + 2 \sqrt{1-\frac x2} \right)^2 & \le \left(1+2^2+2^2 \right) \left(x + 1-\frac x2 + 1 - \frac x2 \right) = 18 \\ \implies \sqrt x + 4 \sqrt{1-\frac x2} & \le \sqrt {18} \approx \boxed{4.243} \end{aligned}$

Equality occurs when $x = \dfrac 29$ .

$x=2\sin^2\theta.~\left(0\le \theta \le \dfrac{\pi}{2}\right)$

$\sqrt{x}+4\sqrt{1-\dfrac{x}{2}} = \sqrt{2}\sin\theta+4\cos\theta = \sqrt{18}\sin(\theta+k).$

The maximum is $\boxed{3\sqrt{2}}.$