To find the saddle points, local minima, and local maxima for the function

$f(x, y) = x^3 + y^3 + 21x^2 - 21y^2 - 1,$

we need to perform the following steps:

1. Calculate the partial derivatives:

We start by finding the first partial derivatives of $f(x, y)$ with respect to $x$ and $y$:

$f_x(x, y) = \frac{\partial}{\partial x} \left(x^3 + y^3 + 21x^2 - 21y^2 - 1\right) = 3x^2 + 42x.$

$f_y(x, y) = \frac{\partial}{\partial y} \left(x^3 + y^3 + 21x^2 - 21y^2 - 1\right) = 3y^2 - 42y.$

2. Set the partial derivatives to zero to find critical points:

Setting $f_x(x, y) = 0$ and $f_y(x, y) = 0$, we get two equations:

1. $3x^2 + 42x = 0$ simplifies to $x(x + 14) = 0$, so $x = 0$ or $x = -14$.
2. $3y^2 - 42y = 0$ simplifies to $y(y - 14) = 0$, so $y = 0$ or $y = 14$.

Thus, the critical points are:

• $(0, 0)$
• $(0, 14)$
• $(-14, 0)$
• $(-14, 14)$

3. Classify the critical points:

To determine whether these points are saddle points, local minima, or local maxima, we use the second derivative test by finding the second partial derivatives and the Hessian determinant $D = f_{xx} f_{yy} - (f_{xy})^2$.

1. Second Partial Derivatives:

   $f_{xx} = \frac{\partial^2 f}{\partial x^2} = 6x + 42,$ $f_{yy} = \frac{\partial^2 f}{\partial y^2} = 6y - 42,$ $f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 0.$

2. Hessian Determinant:

   $D = f_{xx} f_{yy} - (f_{xy})^2 = (6x + 42)(6y - 42).$

Now, we evaluate $D$ and $f_{xx}$ at each critical point to classify them:

• At $(0, 0)$:

  • $f_{xx} = 42$
  • $f_{yy} = -42$
  • $D = 42 \cdot (-42) = -1764$ (negative, so $(0, 0)$ is a saddle point).

• At $(0, 14)$:

  • $f_{xx} = 42$
  • $f_{yy} = 42$
  • $D = 42 \cdot 42 = 1764$ (positive, and $f_{xx} > 0$, so $(0, 14)$ is a local minimum).
  • The value of $f(0, 14) = 0^3 + 14^3 + 21 \cdot 0^2 - 21 \cdot 14^2 - 1 = 2744 - 4116 - 1 = -1373$.

• At $(-14, 0)$:

  • $f_{xx} = -42$
  • $f_{yy} = -42$
  • $D = (-42) \cdot (-42) = 1764$ (positive, and $f_{xx} < 0$, so $(-14, 0)$ is a local maximum).
  • The value of $f(-14, 0) = (-14)^3 + 0^3 + 21 \cdot (-14)^2 - 21 \cdot 0^2 - 1 = -2744 + 4116 - 1 = 1371$.

• At $(-14, 14)$:

  • $f_{xx} = -42$
  • $f_{yy} = 42$
  • $D = (-42) \cdot 42 = -1764$ (negative, so $(-14, 14)$ is a saddle point).

Summary:

• Saddle Points: $(0, 0)$ and $(-14, 14)$
• Local Minimum: Occurs at $(0, 14)$ with value $-1373$
• Local Maximum: Occurs at $(-14, 0)$ with value $1371$

Would you like further details on any part of this solution, or do you have any questions?

Here are some related questions that might help deepen your understanding:

1. How would you find the critical points for a similar function with different powers?
2. Can you describe the significance of the Hessian determinant in classifying critical points?
3. How would the process change if there was a mixed term like $xy$ in the function?
4. What would happen to the classification if $D = 0$ at a critical point?
5. How can this approach be extended to functions of more than two variables?

Tip: When classifying critical points, always remember to check both the Hessian determinant and the second partial derivatives' signs to distinguish between minima, maxima, and saddle points.