Basic Inequality

Using Cauchy-Schwartz inequality, we have:

$\begin{aligned} (x+y)^2 & \le (1+1)(x^2+y^2) = 36 \quad \quad \small \color{#3D99F6}{x^2+y^2 = 18} \\ \Rightarrow x+y & \le \boxed{6} \end{aligned}$

By AM-GM Inequality,

$\frac{x^2+y^2}{2} ≥ \sqrt{x^2y^2} \\ 9≥ xy$

So maximum value, $x+y=\sqrt{x^2+y^2+2xy}=\sqrt{36}=6$

Can also be solved by considering a point on the circle x^2+y^2=18 in first quadrant which gives x+y=3root2(sint+cost) giving max value=6

i solved with c-s. but i am going to use derivatives here just for the sake of variety. since x is positive: $x=\sqrt{18-y^2}$ the required $f(y)=y+x=y+\sqrt{18-y^2}$ for the max(or min) to be achieved, f'(y)=0. $f'(y)=1+(-y)(\sqrt{18-y^2})^{-1}=0$ $\sqrt{18-y^2}=y$ $y^2=9\Longrightarrow y=3$ the max is $f(3)=6$ we check this indeed is max by substituting say y=1.

This is a straightforward application of a two variable case of QA inequality . For proof , see the first example in the wiki. Substitute required values and we get $x+y \leq 6$ .

We can use the QM-AM inequality which states that $\sqrt {\dfrac {x^2+y^2}{2}}\ge \dfrac {x+y}{2}.$ Now it's given that $x^2+y^2=18$ .Substituting this in this inequality,we get $\sqrt {\dfrac {18}{2}}\ge\dfrac {x+y}{2}\therefore 6\ge x+y.$

(X^2+y^2)^1/2=(2×3×3)^1/2

×+y=2×3 =6

$x^2 + y^2 =18$ represents a circle of radius equal to $\sqrt{18}$ and let value of $x+y = k$ the for maximum value the line $x+y=k$ should be tangent to circle so its perpendicular distance from origin is equal to the radius of circle i.e $\sqrt{18}$ which on solving gives value of k as $|0 + 0 - k|\over \sqrt{1^2 +1^2}$ $=\sqrt {18}$ therefore we get $k= \boxed 6$