To classify the critical points of the function $f(x, y) = x^3 - y^2 + 4xy$ using the second derivative test, we need to follow these steps:

Step 1: Find the first partial derivatives

We begin by finding the partial derivatives of $f(x, y)$ with respect to $x$ and $y$.

1. Partial derivative with respect to $x$:

$f_x = \frac{\partial}{\partial x}(x^3 - y^2 + 4xy) = 3x^2 + 4y$

2. Partial derivative with respect to $y$:

$f_y = \frac{\partial}{\partial y}(x^3 - y^2 + 4xy) = -2y + 4x$

Step 2: Find the critical points

Critical points occur where both partial derivatives are equal to zero. Thus, we need to solve the system of equations:

$f_x = 0 \quad \text{and} \quad f_y = 0$

That is, solve:

$3x^2 + 4y = 0 \quad \text{and} \quad -2y + 4x = 0$

Step 3: Solve the system of equations

Let's solve this system.

From $-2y + 4x = 0$, we get:

$y = 2x$

Substitute $y = 2x$ into $3x^2 + 4y = 0$:

$3x^2 + 4(2x) = 0$ $3x^2 + 8x = 0$ $x(3x + 8) = 0$

This gives two solutions for $x$:

$x = 0 \quad \text{or} \quad x = -\frac{8}{3}$

Now, using $y = 2x$, the corresponding values for $y$ are:

• When $x = 0$, $y = 2(0) = 0$
• When $x = -\frac{8}{3}$, $y = 2\left(-\frac{8}{3}\right) = -\frac{16}{3}$

Thus, the critical points are $(0, 0)$ and $\left( -\frac{8}{3}, -\frac{16}{3} \right)$.

Step 4: Calculate the second partial derivatives

Next, we compute the second partial derivatives of $f(x, y)$.

1. Second partial derivative with respect to $x$:

$f_{xx} = \frac{\partial}{\partial x}(3x^2 + 4y) = 6x$

2. Second partial derivative with respect to $y$:

$f_{yy} = \frac{\partial}{\partial y}(-2y + 4x) = -2$

3. Mixed partial derivative $f_{xy}$:

$f_{xy} = \frac{\partial}{\partial y}(3x^2 + 4y) = 4$

Step 5: Apply the second derivative test

For each critical point $(x_0, y_0)$, we compute the discriminant $D$ using the formula:

$D = f_{xx}(x_0, y_0) f_{yy}(x_0, y_0) - (f_{xy}(x_0, y_0))^2$

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

• $f_{xx}(0, 0) = 6(0) = 0$
• $f_{yy}(0, 0) = -2$
• $f_{xy}(0, 0) = 4$

The discriminant is:

$D = (0)(-2) - (4)^2 = -16$

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

For the critical point $\left( -\frac{8}{3}, -\frac{16}{3} \right)$:

• $f_{xx}\left( -\frac{8}{3}, -\frac{16}{3} \right) = 6\left(-\frac{8}{3}\right) = -16$
• $f_{yy}\left( -\frac{8}{3}, -\frac{16}{3} \right) = -2$
• $f_{xy}\left( -\frac{8}{3}, -\frac{16}{3} \right) = 4$

The discriminant is:

$D = (-16)(-2) - (4)^2 = 32 - 16 = 16$

Since $D > 0$ and $f_{xx} = -16 < 0$, the point $\left( -\frac{8}{3}, -\frac{16}{3} \right)$ is a local maximum.

Summary:

• The critical point $(0, 0)$ is a saddle point.
• The critical point $\left( -\frac{8}{3}, -\frac{16}{3} \right)$ is a local maximum.

Would you like further details or clarifications on any part of the process? Here's a list of related questions you might find interesting:

1. How do second partial derivatives help classify critical points?
2. What are the conditions for using the second derivative test?
3. What would happen if $D = 0$ in the second derivative test?
4. How do you classify critical points if $D > 0$ but $f_{xx} > 0$?
5. What is the difference between saddle points and local extrema?

Tip: For functions of two variables, always check the signs of the discriminant $D$ and $f_{xx}$ to determine the nature of the critical points.