To find the critical points of the function $f(x, y)$, we need to find the values of $x$ and $y$ where both partial derivatives are equal to zero.

The partial derivatives are given as:

$f_x = 3x^2 - 24$ $f_y = 3y + 6$

Step 1: Solve $f_x = 0$

$3x^2 - 24 = 0$ $3x^2 = 24$ $x^2 = \frac{24}{3} = 8$ $x = \pm \sqrt{8} = \pm 2\sqrt{2}$

Step 2: Solve $f_y = 0$

$3y + 6 = 0$ $3y = -6$ $y = -2$

Step 3: Find the critical points

The critical points occur where both partial derivatives are zero. From the solutions above:

• $x = 2\sqrt{2}, y = -2$
• $x = -2\sqrt{2}, y = -2$

Critical Points:

The critical points of the function $f(x, y)$ are $(2\sqrt{2}, -2)$ and $(-2\sqrt{2}, -2)$.

Would you like to explore the nature of these critical points (i.e., whether they are maxima, minima, or saddle points)? If you have any other questions or need details, feel free to ask!

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Related Questions:

1. How do you determine if a critical point is a maximum, minimum, or saddle point?
2. What is the significance of a critical point in multivariable calculus?
3. How do you calculate second-order partial derivatives?
4. What is the Hessian matrix, and how is it used in classifying critical points?
5. Can a function have more than one critical point at the same location?
6. What happens if the partial derivatives do not equal zero at any point?
7. How would you visualize a function like $f(x, y)$ on a 3D graph?
8. What are some common applications of finding critical points in real-world scenarios?

Tip: When solving for critical points, always remember to check the second derivative test to classify the points.