Classify the stationary points:f(x,y) = 2x³ -6x² -3y² -6xy 

To classify the stationary points of the function \( f(x, y) = 2x^3 - 6x^2 - 3y^2 - 6xy \), we first need to find the stationary points by calculating the first partial derivatives and setting them to zero.

### Step 1: Find the first partial derivatives

1. **Partial derivative with respect to \( x \)**:
   \[
   f_x = \frac{\partial f}{\partial x} = 6x^2 - 12x - 6y
   \]

2. **Partial derivative with respect to \( y \)**:
   \[
   f_y = \frac{\partial f}{\partial y} = -6y - 6x
   \]

### Step 2: Set the partial derivatives to zero

Set \( f_x = 0 \) and \( f_y = 0 \):
1. \( 6x^2 - 12x - 6y = 0 \)  (1)
2. \( -6y - 6x = 0 \)  (2)

From equation (2), we get:
\[
y = -x
\]

### Step 3: Substitute \( y = -x \) into equation (1)

Substituting into equation (1):
\[
6x^2 - 12x - 6(-x) = 0
\]
\[
6x^2 - 12x + 6x = 0
\]
\[
6x^2 - 6x = 0
\]
\[
6x(x - 1) = 0
\]
This gives us the solutions:
\[
x = 0 \quad \text{or} \quad x = 1
\]

Now substituting back to find \( y \):
- For \( x = 0 \): \( y = -0 = 0 \) → Stationary point \( (0, 0) \)
- For \( x = 1 \): \( y = -1 \) → Stationary point \( (1, -1) \)

### Step 4: Classify the stationary points

We use the second derivative test. First, we calculate the second partial derivatives:

1. **Second partial derivative with respect to \( x \)**:
   \[
   f_{xx} = \frac{\partial^2 f}{\partial x^2} = 12x - 12
   \]

2. **Second partial derivative with respect to \( y \)**:
   \[
   f_{yy} = \frac{\partial^2 f}{\partial y^2} = -6
   \]

3. **Mixed partial derivative**:
   \[
   f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = -6
   \]

Now we can evaluate these at the stationary points:

#### At \( (0, 0) \):
\[
f_{xx}(0, 0) = 12(0) - 12 = -12
\]
\[
f_{yy}(0, 0) = -6
\]
\[
f_{xy}(0, 0) = -6
\]
Calculating the determinant:
\[
D = f_{xx}(0, 0) f_{yy}(0, 0) - (f_{xy}(0, 0))^2 = (-12)(-6) - (-6)^2 = 72 - 36 = 36 > 0
\]
Since \( f_{xx}(0, 0) < 0 \), \( (0, 0) \) is a local maximum.

#### At \( (1, -1) \):
\[
f_{xx}(1, -1) = 12(1) - 12 = 0
\]
\[
f_{yy}(1, -1) = -6
\]
\[
f_{xy}(1, -1) = -6
\]
Calculating the determinant:
\[
D = f_{xx}(1, -1) f_{yy}(1, -1) - (f_{xy}(1, -1))^2 = (0)(-6) - (-6)^2 = 0 - 36 = -36 < 0
\]
Since \( D < 0 \), \( (1, -1) \) is a saddle point.

### Summary of stationary points:
- \( (0, 0) \): Local maximum
- \( (1, -1) \): Saddle point

If you have any questions or need further details, feel free to ask!

### Further Exploration:
1. What are the implications of local maxima and saddle points in optimization?
2. How would you visualize the function to understand its behavior around the stationary points?
3. Can you find any global maxima or minima for this function?
4. How does the presence of a saddle point affect the overall shape of the graph?
5. What are some practical applications of finding stationary points in real-world scenarios?

**Tip:** Always verify second derivatives carefully, as they are crucial for classification.

Classify the stationary points: f(x,y) = 2x³ -6x² -3y² -6xy

Explore how to classify stationary points for the function f(x, y) = 2x³ - 6x² - 3y² - 6xy. This solution details the calculation of partial derivatives, stationary points, and the application of the second derivative test, identifying local maxima and saddle points.

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