To solve this problem, we will follow these steps:

1. Find the critical points by setting the partial derivatives with respect to $x$ and $y$ equal to zero.
2. Use the second derivative test to classify these critical points as local maxima, local minima, or saddle points.
3. Verify the solutions graphically by plotting the level curves and identifying the behavior of the function.

Step 1: Find the Critical Points

Let $f(x, y) = \sin(x) + \sin(y) + \cos(x + y) + 9$.

Partial Derivatives

Compute the partial derivatives with respect to $x$ and $y$:

$f_x = \cos(x) - \sin(x + y)$ $f_y = \cos(y) - \sin(x + y)$

Set these partial derivatives equal to zero to find the critical points:

$\cos(x) - \sin(x + y) = 0$ $\cos(y) - \sin(x + y) = 0$

Subtract the two equations:

$\cos(x) - \cos(y) = 0 \implies \cos(x) = \cos(y)$

This implies that $x = y$ within the given domain $0 \leq x \leq \frac{\pi}{4}$, $0 \leq y \leq \frac{\pi}{4}$.

Solving for $x$ and $y$

With $x = y$, we substitute back into $f_x = 0$:

$\cos(x) - \sin(2x) = 0$

Using the double-angle identity, $\sin(2x) = 2\sin(x)\cos(x)$, we get:

$\cos(x) - 2\sin(x)\cos(x) = 0 \implies \cos(x)(1 - 2\sin(x)) = 0$

Thus, $\cos(x) = 0$ or $\sin(x) = \frac{1}{2}$.

• $\cos(x) = 0$ does not yield any solutions in the interval $0 \leq x \leq \frac{\pi}{4}$.
• $\sin(x) = \frac{1}{2}$ gives $x = \frac{\pi}{6}$. Thus, $x = y = \frac{\pi}{6}$.

The critical point is $\left(\frac{\pi}{6}, \frac{\pi}{6}\right)$.

Step 2: Use the Second Derivative Test

Compute the second partial derivatives:

$f_{xx} = -\sin(x) - \cos(x + y)$ $f_{yy} = -\sin(y) - \cos(x + y)$ $f_{xy} = f_{yx} = -\cos(x + y)$

The discriminant $D$ is given by:

$D = f_{xx} f_{yy} - (f_{xy})^2$

Evaluate $D$ at the critical point $\left( \frac{\pi}{6}, \frac{\pi}{6} \right)$:

$f_{xx} = -\sin\left(\frac{\pi}{6}\right) - \cos\left(\frac{\pi}{3}\right) = -\frac{1}{2} - \frac{1}{2} = -1$ $f_{yy} = -\sin\left(\frac{\pi}{6}\right) - \cos\left(\frac{\pi}{3}\right) = -1$ $f_{xy} = -\cos\left(\frac{\pi}{3}\right) = -\frac{1}{2}$

Thus, the discriminant is:

$D = (-1)(-1) - \left(-\frac{1}{2}\right)^2 = 1 - \frac{1}{4} = \frac{3}{4}$

Since $D > 0$ and $f_{xx} < 0$, the critical point $\left( \frac{\pi}{6}, \frac{\pi}{6} \right)$ is a local maximum.

Step 3: Graph and Level Curves

The level curves can be plotted for $f(x, y)$, showing the contours of the function over the given domain to confirm the local maximum at $\left(\frac{\pi}{6}, \frac{\pi}{6}\right)$. The graphical representation will reveal whether there are any saddle points or other significant features within the domain.

Summary of Results

The function has a local maximum at:

$\left(\frac{\pi}{6}, \frac{\pi}{6}\right), \quad f\left(\frac{\pi}{6}, \frac{\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right) + \sin\left(\frac{\pi}{6}\right) + \cos\left(\frac{\pi}{3}\right) + 9 = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + 9 = 10.5$

There are no other local maxima, minima, or saddle points in the specified domain.

Would you like me to generate the graph and level curves for better visualization?