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To solve this problem, we’ll use the method of **Lagrange multipliers** to find the maximum and minimum values of the function \( f(x, y) = x^2 - y^2 \) subject to the constraint \( g(x, y) = x^2 + y^2 - 64 = 0 \).

### Step 1: Set up the Lagrange function
The method of Lagrange multipliers involves defining a new function \( \mathcal{L}(x, y, \lambda) \):
\[
\mathcal{L}(x, y, \lambda) = f(x, y) - \lambda \cdot g(x, y)
\]
So in this case:
\[
\mathcal{L}(x, y, \lambda) = (x^2 - y^2) - \lambda (x^2 + y^2 - 64)
\]

### Step 2: Compute the partial derivatives
We take partial derivatives of \( \mathcal{L} \) with respect to \( x \), \( y \), and \( \lambda \), and set them equal to zero.

1. **Partial derivative with respect to \( x \):**
   \[
   \frac{\partial \mathcal{L}}{\partial x} = 2x - \lambda \cdot 2x = 0
   \]
   Simplifying, we get:
   \[
   x(1 - \lambda) = 0
   \]

2. **Partial derivative with respect to \( y \):**
   \[
   \frac{\partial \mathcal{L}}{\partial y} = -2y - \lambda \cdot 2y = 0
   \]
   Simplifying, we get:
   \[
   y(-1 - \lambda) = 0
   \]

3. **Partial derivative with respect to \( \lambda \):**
   \[
   \frac{\partial \mathcal{L}}{\partial \lambda} = -(x^2 + y^2 - 64) = 0
   \]
   which gives us the constraint:
   \[
   x^2 + y^2 = 64
   \]

### Step 3: Solve the equations
From the equations \( x(1 - \lambda) = 0 \) and \( y(-1 - \lambda) = 0 \), we have two cases for \( \lambda \):

1. **Case 1: \( x = 0 \)**

   If \( x = 0 \), then \( x^2 + y^2 = 64 \) implies \( y^2 = 64 \), so \( y = \pm 8 \).
   
   - For \( (x, y) = (0, 8) \), we get \( f(0, 8) = 0 - 8^2 = -64 \).
   - For \( (x, y) = (0, -8) \), we get \( f(0, -8) = 0 - (-8)^2 = -64 \).

2. **Case 2: \( y = 0 \)**

   If \( y = 0 \), then \( x^2 + y^2 = 64 \) implies \( x^2 = 64 \), so \( x = \pm 8 \).
   
   - For \( (x, y) = (8, 0) \), we get \( f(8, 0) = 8^2 - 0 = 64 \).
   - For \( (x, y) = (-8, 0) \), we get \( f(-8, 0) = (-8)^2 - 0 = 64 \).

### Conclusion
The maximum value of \( f(x, y) \) is \( 64 \), which occurs at \( (x, y) = (8, 0) \) and \( (-8, 0) \).

The minimum value of \( f(x, y) \) is \( -64 \), which occurs at \( (x, y) = (0, 8) \) and \( (0, -8) \).

### Final Answer
- **Maximum value:** \( 64 \)
- **Minimum value:** \( -64 \)

Would you like a more detailed breakdown, or do you have any questions?

### Related Questions
1. How would the solution change if the constraint was different, like \( x^2 + y^2 = 100 \)?
2. Can you explain the role of Lagrange multipliers in optimization with constraints?
3. How would you solve this problem if \( f(x, y) \) was different, for example \( f(x, y) = x^2 + y^2 \)?
4. What would be the solution if the constraint were an inequality rather than an equation?
5. How do you interpret the physical meaning of this problem if \( f(x, y) \) represents height and \( x, y \) are coordinates?

### Tip
Lagrange multipliers are especially useful in physics and engineering, where constraints like energy conservation often require optimization with multiple variables.

This extreme value problem has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to find the extreme values of the function subject to the given constraint: f(x, y) = x^2 - y^2, x^2 + y^2 = 64

Learn to solve an optimization problem using Lagrange multipliers to find the maximum and minimum values of f(x, y) = x^2 - y^2 under the constraint x^2 + y^2 = 64. This solution includes key calculus concepts and step-by-step calculations suitable for advanced high school students.

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