No using calculator

Let the sum be $\sqrt{x} + \sqrt{y}$ , and we note that $x+y=16$ . Since $\sqrt{x}, \sqrt{y} > 0$ , we can use Cauchy-Schwarz inequality.

$\begin{aligned} (\sqrt{x} + \sqrt{y})^2 & \le 2 (x+y) = 32 \\ \Rightarrow \sqrt{x} + \sqrt{y} & \le 2 \sqrt{8} \end{aligned}$

Equality or maximum condition occurs when $x=y = 8$ . The maximum sum is therefore $\boxed{(\text{A})} \space \sqrt{8} + \sqrt{8}$ .

We will prove that for non-negative real number $x$ , $\sqrt{8-x}+\sqrt{8+x}$ obtains maximum value when $x=0$ .

Indeed, we have:

$(\sqrt{8-x}+\sqrt{8+x})^2=16+2\sqrt{64-x^2}\le 16+2\sqrt{64}=32$

Thus, $\sqrt{8-x}+\sqrt{8+x}\le 4\sqrt{2}$ and the equality holds iff $x=0$ .

So, the largest number is $\boxed{A}$ .

We noticed that8+8=7+9=6+10=5+11=4+12=16 Square each of that we will have A=16+2sqrt(64) B=16+2sqrt(63) C=16+2sqrt(60) D=16+2sqrt(55) E=16+2sqrt(48) Hence A is the largest

Root 8"= 2root2 =3.141×2=6.28 Rest are betn 5&6

The derivative of root x is monotonically decreasing. That is, as x increases, it increases by decreasing amounts. So when you subtract from 8, your loss is much larger than the gain accrued by adding to 8. This helps conclude that a) must be the answer.