Solution Verification: Find the Global Max and Min of $z=x+y$, in $D=\{x\ge 0, y\ge 0, y\le 3x, x^2+y^2\le 25 \}$ 
$z=x+y$
$D=\{x\ge 0, y\ge 0, y\le 3x, x^2+y^2\le 25 \}$

My attempt:
Plan: 
Find inner critical points. 
Find maxmin of border curves. 
Find intersecting points of border curves. 
Find values of all points and decide global maximum and minimum.

Inner critical points:

First of all, I checked the inner points of the domain by taking $z_x,z_y=0$. And obviously there's no critical points there since $z_x=z_y=1$. 
Borders or domain curves (not sure what it's called):

Next step, is finding the maximum and minimum on the domain curves (Or the borders), 
$x^2+y^2=25 \Longrightarrow x = \sqrt{25-y^2}$, Substituting that into my function: 
$z = \sqrt{25-y^2}+y \Longrightarrow z'=\frac{-2y}{2\sqrt{25-y^2}}+1=0$ (To find critical points). 
$-y+\sqrt{25-y^2}=0 \Longrightarrow y^2=25-y^2 \Longrightarrow y^2 =\frac{25}{2} \Longrightarrow y=\frac{5}{\sqrt{2}}$ (I took positive value only because $y \ge 0$). 
The point $(x,\frac{5}{\sqrt{2}})$ must be on the circle, so $x^2+\frac{25}{2}=25 \Longrightarrow x=\frac{5}{\sqrt{2}}$ ($x \ge 0$). 
So I've found a point $(\frac{5}{\sqrt{2}},\frac{5}{\sqrt{2}})$
Next, $y=0 \Longrightarrow z=x \Longrightarrow z_x=1$ There's no critical points. (same for $x=0$). 
Next, $y=3x\Longrightarrow z=4x \Longrightarrow z_x = 4$ (no critical points too). 
Intersecting points of border curves:

Now my next step is to check points where $x^2+y^2=25$ and $y=3x$ and $y=0$ and $x=0$ meet, 
I can see points $(0,0)$, $(0,5)$, now I need to find the intersection of the circle and $y=3x$. 
$(x^2 + 9x^2)=25 \Longrightarrow 10x^2=25 \Longrightarrow x=\frac{5}{\sqrt{10}}$ 
Substituting in the $y=3x \Longrightarrow y=\frac{15}{\sqrt{10}}$ And so my last point is $(\frac{5}{\sqrt{10}}, \frac{15}{\sqrt{10}})$ 
Calculating the value of the function in the points I've found:

My next and last step, is to calculate the value of $z=f(x,y)$ in all the points I have found: 
$f(\frac{5}{\sqrt{2}},\frac{5}{\sqrt{2}})= \frac{10}{\sqrt{2}} = 7.071...$ 
$f(0,0)=0$ 
$f(0,5)=5$ 
$f(\frac{5}{\sqrt{10}}, \frac{15}{\sqrt{10}})= \frac{20}{\sqrt{10}}=6.324...$
Final Answer:

And so, my global maximum is $(\frac{5}{\sqrt{2}},\frac{5}{\sqrt{2}})$, Global minimum is $(0,0)$. 
I would love to hear feedback about my solution and to know if I missed some points, checked some unnecessary points or some of my work flow doesn't seem okay. Thanks in advance!
 A: Geometrically, we are bound by circle $x^2 + y^2 = 25 \ $ between $x$-axis and line $y = 3x$ in the first quadrant.
In polar coordinates, $x = r \cos\theta, y = r \sin\theta$.
Constraint is given as $x^2 + y^2 \leq 25, x,y \geq 0 \implies r \leq 5, 0 \leq \theta \leq \frac{\pi}{2}$
But we are also bound by $y = 3x, \tan \theta = 3 \implies 0 \leq \theta \leq \arctan(3)$
So we need to find minimum and maximum of $z = x + y = r ( \cos\theta + \sin\theta) \ $ in region $0 \leq r \leq 5, 0 \leq \theta \leq \arctan(3)$
We know $ \ 0 \lt \cos\theta + \sin\theta \leq \sqrt2 \ $ for $0 \leq \theta \leq \frac{\pi}{2}$.
We also know (or can quickly take derivative to find it), maximum of $\sqrt2 \ $ occurs at $ \theta = \frac{\pi}{4} \lt \arctan (3)$, which is in our region.
So it is clear that maximum of function $z$ occurs at $r = 5$ and $\theta = \frac{\pi}{4}$ which is point $(\frac{5}{\sqrt2}, \frac{5}{\sqrt2})$.
As $\cos\theta + \sin\theta \gt 0$ and $r \geq 0$, it is easy to see that $z = r (\cos\theta + \sin\theta)$ will have minimum at $r = 0$, which is point $(0, 0)$.
