Answer
There are two types of critical points:
1. Type 1 critical points: $\left( {x,y} \right) = \left( {\frac{\pi }{3} - \frac{1}{2} + 2n\pi ,\frac{1}{2}} \right)$ for $n = 0,1,2,...$
These are are local maxima.
2. Type 2 critical points: $\left( {x,y} \right) = \left( { - \frac{\pi }{3} - \frac{1}{2} + 2n\pi ,\frac{1}{2}} \right)$ for $n = 0,1,2,...$
These are saddle points.
As an illustration, please see the figure attached.
Work Step by Step
We have $f\left( {x,y} \right) = \sin \left( {x + y} \right) - \frac{1}{2}\left( {x + {y^2}} \right)$.
The partial derivatives are
${f_x} = \cos \left( {x + y} \right) - \frac{1}{2}$, ${\ \ \ }$ ${f_y} = \cos \left( {x + y} \right) - y$
To find the critical points we solve the equations ${f_x} = 0$ and ${f_y} = 0$:
${f_x} = \cos \left( {x + y} \right) - \frac{1}{2} = 0$
$\cos \left( {x + y} \right) = \frac{1}{2}$
${f_y} = \cos \left( {x + y} \right) - y = 0$
$y = \cos \left( {x + y} \right)$
From these two equations we obtain $y = \frac{1}{2}$.
Substituting $y = \frac{1}{2}$ in $\cos \left( {x + y} \right) = \frac{1}{2}$ gives
$\cos \left( {x + \frac{1}{2}} \right) = \frac{1}{2}$
The solutions are $x + \frac{1}{2} = \pm \frac{\pi }{3} + 2n\pi $ for $n = 0,1,2,...$
So, there are two types of critical points:
Type 1: ${\ \ \ \ }$ $\left( {x,y} \right) = \left( {\frac{\pi }{3} - \frac{1}{2} + 2n\pi ,\frac{1}{2}} \right)$
Type 2: ${\ \ \ \ }$ $\left( {x,y} \right) = \left( { - \frac{\pi }{3} - \frac{1}{2} + 2n\pi ,\frac{1}{2}} \right)$
for $n = 0,1,2,...$
The Second Derivative Test:
We evaluate the second partial derivatives:
${f_{xx}} = - \sin \left( {x + y} \right)$, ${\ \ \ }$ ${f_{yy}} = - \sin \left( {x + y} \right) - 1$, ${\ \ \ }$ ${f_{xy}} = - \sin \left( {x + y} \right)$
The discriminant:
$D = {f_{xx}}{f_{yy}} - {f_{xy}}^2$
$ = {\sin ^2}\left( {x + y} \right) + \sin \left( {x + y} \right) - {\sin ^2}\left( {x + y} \right)$
$ = \sin \left( {x + y} \right)$
We evaluate the second partial derivatives at the critical points:
Type 1 critical points: $\left( {x,y} \right) = \left( {\frac{\pi }{3} - \frac{1}{2} + 2n\pi ,\frac{1}{2}} \right)$ for $n = 0,1,2,...$
${f_{xx}} = - \sin \left( {x + y} \right)$
$ = - \sin \left( {\frac{\pi }{3} + 2n\pi } \right)$
$ = - \sin \left( {\frac{\pi }{3}} \right)\cos \left( {2n\pi } \right) - \cos \left( {\frac{\pi }{3}} \right)\sin \left( {2n\pi } \right)$
$ = - \sin \left( {\frac{\pi }{3}} \right)$
$ = - \frac{{\sqrt 3 }}{2}$
So, $D = \sin \left( {\frac{\pi }{3} + 2n\pi } \right) = \frac{{\sqrt 3 }}{2}$.
Since $D>0$ and ${f_{xx}} < 0$, by the Second Derivative Test we conclude that type 1 critical points are local maxima.
Type 2 critical points: $\left( {x,y} \right) = \left( { - \frac{\pi }{3} - \frac{1}{2} + 2n\pi ,\frac{1}{2}} \right)$ for $n = 0,1,2,...$
${f_{xx}} = - \sin \left( {x + y} \right)$
$ = - \sin \left( { - \frac{\pi }{3} + 2n\pi } \right)$
$ = - \sin \left( { - \frac{\pi }{3}} \right)\cos \left( {2n\pi } \right) - \cos \left( { - \frac{\pi }{3}} \right)\sin \left( {2n\pi } \right)$
$ = - \sin \left( { - \frac{\pi }{3}} \right)$
$ = \frac{{\sqrt 3 }}{2}$
So, $D = \sin \left( { - \frac{\pi }{3} + 2n\pi } \right) = - \frac{{\sqrt 3 }}{2}$.
Since $D<0$, by the Second Derivative Test we conclude that type 2 critical points are saddle points.
