To solve for the local maxima, minima, and saddle points of the function:

$f(x, y) = xy - 3x - 3y - x^2 - y^2$

We proceed as follows:

Step 1: Partial Derivatives

First, compute the first partial derivatives of $f(x, y)$ with respect to $x$ and $y$:

$f_x = \frac{\partial f}{\partial x} = y - 3 - 2x$ $f_y = \frac{\partial f}{\partial y} = x - 3 - 2y$

Step 2: Critical Points

To find critical points, set $f_x = 0$ and $f_y = 0$:

1. From $f_x = 0$:
   $y - 3 - 2x = 0 \quad \Rightarrow \quad y = 2x + 3$

2. From $f_y = 0$:
   $x - 3 - 2y = 0 \quad \Rightarrow \quad x = 2y + 3$

Substitute $y = 2x + 3$ into $x = 2y + 3$:

$x = 2(2x + 3) + 3$ $x = 4x + 6 + 3$ $x - 4x = 9 \quad \Rightarrow \quad -3x = 9 \quad \Rightarrow \quad x = -3$

Now substitute $x = -3$ into $y = 2x + 3$:

$y = 2(-3) + 3 = -6 + 3 = -3$

Thus, the critical point is:

$(x, y) = (-3, -3)$

Step 3: Second Derivative Test

To classify the critical point, compute the second partial derivatives:

$f_{xx} = \frac{\partial^2 f}{\partial x^2} = -2, \quad f_{yy} = \frac{\partial^2 f}{\partial y^2} = -2, \quad f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 1$

Now calculate the Hessian determinant $D$:

$D = f_{xx}f_{yy} - (f_{xy})^2$ $D = (-2)(-2) - (1)^2 = 4 - 1 = 3$

• Since $D > 0$ and $f_{xx} < 0$, the critical point $(-3, -3)$ is a local maximum.

Step 4: Results

1. Local maximum value:
   Evaluate $f(x, y)$ at $(-3, -3)$:

   $f(-3, -3) = (-3)(-3) - 3(-3) - 3(-3) - (-3)^2 - (-3)^2$ $f(-3, -3) = 9 + 9 + 9 - 9 - 9 = 9$

   So the local maximum value is 9.

2. Local minimum values:
   None exist because the Hessian test did not indicate minima.

3. Saddle points:
   None exist because $D > 0$, indicating no saddle points.

Final Answers:

• Local maximum value(s): $9$
• Local minimum value(s): DNE
• Saddle point(s): $(x, y) = \text{DNE}$

Do you need further clarification or a breakdown of any step?

Related Questions:

1. How is the Hessian determinant used to classify critical points?
2. What happens when $D = 0$ in the second derivative test?
3. Can this function have global maxima or minima?
4. How do saddle points differ from maxima or minima?
5. How do partial derivatives relate to gradients in multivariable calculus?

Tip:

For multivariable functions, the second derivative test relies on the Hessian determinant to determine whether critical points are maxima, minima, or saddle points. Always check $D$ and $f_{xx}$ carefully!