Find  the  local  maxima,  local  minima  and  saddle  point  (if  any  of  them exists)  of  the  function  g (x, y )  =  x^3 - 12xy + 8y^3

To find the local maxima, local minima, and saddle points of the function \( g(x, y) = x^3 - 12xy + 8y^3 \), we will follow these steps:

1. **Find the first-order partial derivatives** of the function.
2. **Set the first-order partial derivatives to zero** to find the critical points.
3. **Determine the second-order partial derivatives**.
4. **Use the second derivative test** to classify each critical point as a local maximum, local minimum, or saddle point.

### Step 1: First-order Partial Derivatives

The first-order partial derivatives of \( g(x, y) = x^3 - 12xy + 8y^3 \) are:

\[
g_x(x, y) = \frac{\partial g}{\partial x} = 3x^2 - 12y
\]

\[
g_y(x, y) = \frac{\partial g}{\partial y} = -12x + 24y^2
\]

### Step 2: Set the First-order Partial Derivatives to Zero

To find the critical points, we solve the equations:

\[
g_x(x, y) = 0 \implies 3x^2 - 12y = 0
\]

\[
g_y(x, y) = 0 \implies -12x + 24y^2 = 0
\]

#### Solving the Equations

1. From \( g_x(x, y) = 0 \):

\[
3x^2 = 12y \implies x^2 = 4y \implies y = \frac{x^2}{4}
\]

2. Substitute \( y = \frac{x^2}{4} \) into \( g_y(x, y) = 0 \):

\[
-12x + 24\left( \frac{x^2}{4} \right)^2 = 0
\]

\[
-12x + 24 \cdot \frac{x^4}{16} = 0
\]

\[
-12x + \frac{3x^4}{2} = 0
\]

\[
3x^4 = 24x \implies x(x^3 - 8) = 0
\]

So, \( x = 0 \) or \( x^3 = 8 \implies x = 2 \).

#### Finding Corresponding \( y \) Values

- If \( x = 0 \):

\[
y = \frac{0^2}{4} = 0
\]

So, one critical point is \((0, 0)\).

- If \( x = 2 \):

\[
y = \frac{2^2}{4} = 1
\]

So, another critical point is \((2, 1)\).

### Step 3: Second-order Partial Derivatives

Compute the second-order partial derivatives:

\[
g_{xx} = \frac{\partial^2 g}{\partial x^2} = 6x
\]

\[
g_{yy} = \frac{\partial^2 g}{\partial y^2} = 48y
\]

\[
g_{xy} = \frac{\partial^2 g}{\partial x \partial y} = -12
\]

### Step 4: Second Derivative Test

For a function \( g(x, y) \), the second derivative test involves the determinant of the Hessian matrix \( H \):

\[
D(x, y) = g_{xx}(x, y) \cdot g_{yy}(x, y) - (g_{xy}(x, y))^2
\]

1. **At \((0, 0)\):**

\[
g_{xx}(0, 0) = 0, \quad g_{yy}(0, 0) = 0, \quad g_{xy}(0, 0) = -12
\]

\[
D(0, 0) = (0)(0) - (-12)^2 = -144
\]

Since \( D(0, 0) < 0 \), \((0, 0)\) is a **saddle point**.

2. **At \((2, 1)\):**

\[
g_{xx}(2, 1) = 6 \cdot 2 = 12, \quad g_{yy}(2, 1) = 48 \cdot 1 = 48, \quad g_{xy}(2, 1) = -12
\]

\[
D(2, 1) = (12)(48) - (-12)^2 = 576 - 144 = 432
\]

Since \( D(2, 1) > 0 \) and \( g_{xx}(2, 1) > 0 \), \((2, 1)\) is a **local minimum**.

### Conclusion

- **Saddle Point**: \((0, 0)\)
- **Local Minimum**: \((2, 1)\)
- **No Local Maximum** is present.

Would you like more details or have any questions?

### Related Questions

1. How do we find the global extrema of a function with multiple variables?
2. What is the geometric interpretation of a saddle point?
3. Can a function have multiple saddle points? How do we identify them?
4. What are the applications of finding local minima and maxima in real-world scenarios?
5. How does the second derivative test differ for functions of one variable vs. multiple variables?

### Tip

When analyzing functions of multiple variables, always visualize the critical points with the context of the function's surface plot to better understand their nature.

Discover how to find local maxima, minima, and saddle points of the function g(x, y) = x^3 - 12xy + 8y^3 using partial derivatives and the second derivative test. Detailed steps and critical point analysis included.

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