https://r2.erweima.ai/mathbot/image/1725320610305kr6n2z64.jpg

To determine the local minima and maxima of the function \( f(x, y) = x^3 + y^3 - 6xy \), we'll need to follow these steps:

### Step 1: Compute the partial derivatives
First, find the first partial derivatives of \( f(x, y) \) with respect to \( x \) and \( y \):

\[
f_x(x, y) = \frac{\partial}{\partial x}(x^3 + y^3 - 6xy) = 3x^2 - 6y
\]

\[
f_y(x, y) = \frac{\partial}{\partial y}(x^3 + y^3 - 6xy) = 3y^2 - 6x
\]

### Step 2: Set the partial derivatives equal to zero
To find the critical points, we set the partial derivatives equal to zero:

\[
3x^2 - 6y = 0 \quad \text{(1)}
\]
\[
3y^2 - 6x = 0 \quad \text{(2)}
\]

### Step 3: Solve the system of equations
We can solve this system of equations to find the critical points.

From equation (1):
\[
x^2 = 2y \quad \text{(3)}
\]

From equation (2):
\[
y^2 = 2x \quad \text{(4)}
\]

Substituting equation (3) into equation (4):
\[
y^2 = 2\sqrt{2y}
\]
Square both sides to eliminate the square root:
\[
(y^2)^2 = 4y \implies y^4 = 4y
\]
\[
y(y^3 - 4) = 0
\]
So, \( y = 0 \) or \( y^3 = 4 \), which gives \( y = \sqrt[3]{4} \).

If \( y = 0 \), from equation (3), \( x = 0 \).

If \( y = \sqrt[3]{4} \), then substituting back into equation (3), we get:
\[
x^2 = 2\sqrt[3]{4}
\]
\[
x = \pm \sqrt{2\sqrt[3]{4}}
\]

Thus, the critical points are:
- \( (0, 0) \)
- \( \left(\sqrt{2\sqrt[3]{4}}, \sqrt[3]{4}\right) \)
- \( \left(-\sqrt{2\sqrt[3]{4}}, \sqrt[3]{4}\right) \)

### Step 4: Use the second derivative test
To classify these critical points as minima, maxima, or saddle points, we use the second derivative test. Compute the second partial derivatives:

\[
f_{xx} = \frac{\partial}{\partial x}(3x^2 - 6y) = 6x
\]

\[
f_{yy} = \frac{\partial}{\partial y}(3y^2 - 6x) = 6y
\]

\[
f_{xy} = \frac{\partial}{\partial y}(3x^2 - 6y) = -6
\]

### Step 5: Determine the discriminant \( D \)
The discriminant \( D \) is given by:

\[
D = f_{xx}f_{yy} - (f_{xy})^2
\]

Evaluate \( D \) at each critical point:

1. **At \( (0, 0) \):**
   \[
   D = (6 \times 0)(6 \times 0) - (-6)^2 = -36 \quad \text{(saddle point)}
   \]

2. **At \( \left(\sqrt{2\sqrt[3]{4}}, \sqrt[3]{4}\right) \):**
   \[
   D = (6\sqrt{2\sqrt[3]{4}})(6\sqrt[3]{4}) - (-6)^2 > 0
   \]
   Since \( f_{xx} > 0 \) and \( D > 0 \), this point is a **local minimum**.

3. **At \( \left(-\sqrt{2\sqrt[3]{4}}, \sqrt[3]{4}\right) \):**
   \[
   D = (6(-\sqrt{2\sqrt[3]{4}}))(6\sqrt[3]{4}) - (-6)^2 > 0
   \]
   Since \( f_{xx} < 0 \) and \( D > 0 \), this point is a **local maximum**.

### Summary
- **Saddle point** at \( (0, 0) \)
- **Local minimum** at \( \left(\sqrt{2\sqrt[3]{4}}, \sqrt[3]{4}\right) \)
- **Local maximum** at \( \left(-\sqrt{2\sqrt[3]{4}}, \sqrt[3]{4}\right) \)

Would you like further details or have any other questions?

### Follow-up Questions:
1. How do we interpret the second derivative test results?
2. What is the significance of the discriminant \( D \) being negative?
3. How can the nature of critical points influence the shape of the function's graph?
4. Can there be global minima or maxima for this function? Why or why not?
5. How would you proceed if the discriminant was zero?

**Tip:** When dealing with higher-dimensional functions, the discriminant helps classify critical points based on the curvature of the function at those points.

Learn how to determine local minima and maxima of the multivariable function f(x, y) = x^3 + y^3 - 6xy with detailed step-by-step solutions and applications of the second derivative test. Suitable for advanced college-level students and mathematicians.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation