Use a graph and/or level curves to find the local maximum and minimum values and saddle points of the function. Then use calculus to find these values precisely. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)

*f*(*x*, *y*) = sin(*x*) + sin(*y*) + cos(*x* + *y*) + 9,    0 ≤ *x* ≤ 𝜋/4,    0 ≤ *y* ≤ 𝜋/4

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?

Use a graph and/or level curves to find the local maximum and minimum values and saddle points of the function. Then use calculus to find these values precisely. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) f(x, y) = sin(x) + sin(y) + cos(x + y) + 9, 0 ≤ x ≤ π/4, 0 ≤ y ≤ π/4

Use multivariable calculus to solve for local maxima, minima, and saddle points of the function f(x, y) = sin(x) + sin(y) + cos(x + y) + 9 within the domain 0 ≤ x, y ≤ π/4. Learn how to find critical points using partial derivatives and confirm using the second derivative test.

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