2. (14 pts) The two parts below are not related. (a) Find and classify all critical points of $$ f(x, y)=3 x y-\frac{1}{2} y^{2}+2 x^{3}+\frac{9}{2} x^{2} $$. Clearly show your work and reasoning in using the 2nd derivative test.

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To find and classify the critical points of the function $$ f(x, y) = 3xy - \frac{1}{2}y^2 + 2x^3 + \frac{9}{2}x^2 $$, we will follow these steps:

1. **Find the first partial derivatives** $$ f_x $$ and $$ f_y $$.
2. **Set the first partial derivatives to zero** to find critical points.
3. **Find the second partial derivatives** $$ f_{xx} $$, $$ f_{yy} $$, and $$ f_{xy} $$.
4. **Use the second derivative test** to classify the critical points.

### Step 1: Find the first partial derivatives

The first partial derivatives are calculated as follows:

$$
f_x = \frac{\partial f}{\partial x} = 3y + 6x^2 + 9x
$$

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

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

Now we set the first partial derivatives equal to zero:

1. $$ 3y + 6x^2 + 9x = 0 $$  (Equation 1)
2. $$ 3x - y = 0 $$  (Equation 2)

From Equation 2, we can express $$ y $$ in terms of $$ x $$:

$$
y = 3x
$$

Substituting $$ y = 3x $$ into Equation 1:

$$
3(3x) + 6x^2 + 9x = 0
$$

This simplifies to:

$$
9x + 6x^2 + 9x = 0
$$

$$
6x^2 + 18x = 0
$$

Factoring out $$ 6x $$:

$$
6x(x + 3) = 0
$$

This gives us two solutions:

1. $$ x = 0 $$
2. $$ x = -3 $$

Now we can find the corresponding $$ y $$ values:

- For $$ x = 0 $$:
  $$
  y = 3(0) = 0 \quad \Rightarrow \quad (0, 0)
  $$

- For $$ x = -3 $$:
  $$
  y = 3(-3) = -9 \quad \Rightarrow \quad (-3, -9)
  $$

Thus, the critical points are $$ (0, 0) $$ and $$ (-3, -9) $$.

### Step 3: Find the second partial derivatives

Next, we calculate the second partial derivatives:

$$
f_{xx} = \frac{\partial^2 f}{\partial x^2} = 12x + 9
$$

$$
f_{yy} = \frac{\partial^2 f}{\partial y^2} = -1
$$

$$
f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 3
$$

### Step 4: Use the second derivative test

The second derivative test involves calculating the determinant $$ D $$ of the Hessian matrix:

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

Now we evaluate $$ D $$ at each critical point.

#### For the critical point $$ (0, 0) $$:

1. Calculate $$ f_{xx}(0, 0) $$:
   $$
   f_{xx}(0, 0) = 12(0) + 9 = 9
   $$

2. Calculate $$ f_{yy}(0, 0) $$:
   $$
   f_{yy}(0, 0) = -1
   $$

3. Calculate $$ f_{xy}(0, 0) $$:
   $$
   f_{xy}(0, 0) = 3
   $$

4. Calculate $$ D(0, 0) $$:
   $$
   D(0, 0) = 9(-1) - (3)^2 = -9 - 9 = -18
   $$

Since $$ D < 0 $$, the critical point $$ (0, 0) $$ is a saddle point.

#### For the critical point $$ (-3, -9) $$:

1. Calculate $$ f_{xx}(-3, -9) $$:
   $$
   f_{xx}(-3, -9) = 12(-3) + 9 = -36 + 9 = -27
   $$

2. Calculate $$ f_{yy}(-3, -9) $$:
   $$
   f_{yy}(-3, -9) = -1
   $$

3. Calculate $$ f_{xy}(-3, -9) $$:
   $$
   f_{xy}(-3, -9) = 3
   $$

4. Calculate $$ D(-3, -9) $$:
   $$
   D(-3, -9) = (-27)(-1) - (3)^2 = 27 - 9 = 18
   $$

Since $$ D > 0 $$ and $$ f_{xx}(-3, -9) < 0 $$, the critical point $$ (-3, -9) $$ is a local maximum.

### Summary of Critical Points

- The critical point $$ (0, 0) $$ is a saddle point.
- The critical point $$ (-3, -9) $$ is a local maximum.
The critical points of the function $$ f(x, y) = 3xy - \frac{1}{2}y^2 + 2x^3 + \frac{9}{2}x^2 $$ are $$ (0, 0) $$ and $$ (-3, -9) $$. - At $$ (0, 0) $$, the function has a saddle point. - At $$ (-3, -9) $$, the function has a local maximum.

2. (14 pts) The two parts below are not related. (a) Find and classify all critical points of \( f(x, y)=3 x y-\frac{1}{2} y^{2}+2 x^{3}+\frac{9}{2} x^{2} \). Clearly show your work and reasoning in using the 2nd derivative test.

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