To find the critical points of the function $f(x, y) = x^3 + y^3 - 12x^2 - 48y - 3$, we need to follow these steps:

Step 1: Compute the first partial derivatives of $f(x, y)$

The first partial derivatives are:

$f_x = \frac{\partial}{\partial x} \left( x^3 + y^3 - 12x^2 - 48y - 3 \right)$ $f_y = \frac{\partial}{\partial y} \left( x^3 + y^3 - 12x^2 - 48y - 3 \right)$

Let's compute these derivatives:

• For $f_x$, treat $y$ as a constant: $f_x = 3x^2 - 24x$

• For $f_y$, treat $x$ as a constant: $f_y = 3y^2 - 48$

Step 2: Set the first partial derivatives equal to zero to find critical points

Solve the system of equations: $f_x = 3x^2 - 24x = 0$ $f_y = 3y^2 - 48 = 0$

Solve for $x$:

$3x^2 - 24x = 0$ Factor out: $3x(x - 8) = 0$ Thus, $x = 0$ or $x = 8$.

Solve for $y$:

$3y^2 - 48 = 0$ $y^2 = 16$ Thus, $y = 4$ or $y = -4$.

Step 3: List the critical points

From the solutions above, the possible critical points are: $(0, 4), (0, -4), (8, 4), (8, -4)$

Step 4: Classify each critical point using the second derivative test

To classify each critical point, we need the second partial derivatives:

$f_{xx} = \frac{\partial^2 f}{\partial x^2} = 6x - 24$ $f_{yy} = \frac{\partial^2 f}{\partial y^2} = 6y$ f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 0 \quad \text{(since there are no mixed terms involving both x$and$y)}

Now, calculate the discriminant $D$ for each critical point: $D = f_{xx} f_{yy} - (f_{xy})^2 = (6x - 24)(6y) - 0^2 = 6(6x - 24)y$

For $(0, 4)$:

$f_{xx} = 6(0) - 24 = -24, \quad f_{yy} = 6(4) = 24, \quad D = (6(-24))(4) = -576$ Since $D < 0$, $(0, 4)$ is a saddle point.

For $(0, -4)$:

$f_{xx} = 6(0) - 24 = -24, \quad f_{yy} = 6(-4) = -24, \quad D = (6(-24))(-4) = 576$ Since $D > 0$ and $f_{xx} < 0$, $(0, -4)$ is a local maximum.

For $(8, 4)$:

$f_{xx} = 6(8) - 24 = 24, \quad f_{yy} = 6(4) = 24, \quad D = (6(24))(4) = 576$ Since $D > 0$ and $f_{xx} > 0$, $(8, 4)$ is a local minimum.

For $(8, -4)$:

$f_{xx} = 6(8) - 24 = 24, \quad f_{yy} = 6(-4) = -24, \quad D = (6(24))(-4) = -576$ Since $D < 0$, $(8, -4)$ is a saddle point.

Final Classification:

• $(0, 4)$ is a saddle point.
• $(0, -4)$ is a local maximum.
• $(8, 4)$ is a local minimum.
• $(8, -4)$ is a saddle point.

Would you like any further details or clarifications on these results?

---

Here are 5 related questions you could explore:

1. How do second derivative tests help in determining the nature of critical points?
2. What would happen to the classification of critical points if the function had higher-degree terms?
3. How does the presence of mixed partial derivatives (like $f_{xy}$) affect the classification process?
4. How can we apply these techniques in 3D for functions involving $x$, $y$, and $z$?
5. What role does the discriminant $D$ play in the second derivative test?

Tip: When solving for critical points, always check that you’ve correctly calculated the first and second derivatives before proceeding to the classification step.