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To solve this problem, we need to find the extreme values of \( f(x, y) = x - 2y \) subject to the constraint \( g(x, y) = x^2 + y^2 = 20 \) using the method of Lagrange multipliers.

### Step-by-Step Solution

1. **Set up the Lagrange Function**:  
   Define the Lagrange function as:
   \[
   \mathcal{L}(x, y, \lambda) = f(x, y) - \lambda (g(x, y) - 20)
   \]
   So,
   \[
   \mathcal{L}(x, y, \lambda) = x - 2y - \lambda (x^2 + y^2 - 20)
   \]

2. **Compute the Partial Derivatives**:
   We need to find the partial derivatives of \(\mathcal{L}\) with respect to \(x\), \(y\), and \(\lambda\), and set them equal to zero.
   \[
   \frac{\partial \mathcal{L}}{\partial x} = 1 - 2\lambda x = 0
   \]
   \[
   \frac{\partial \mathcal{L}}{\partial y} = -2 - 2\lambda y = 0
   \]
   \[
   \frac{\partial \mathcal{L}}{\partial \lambda} = -(x^2 + y^2 - 20) = 0
   \]

3. **Solve the System of Equations**:
   - From \(\frac{\partial \mathcal{L}}{\partial x} = 0\):  
     \[
     1 = 2\lambda x \Rightarrow \lambda = \frac{1}{2x}
     \]
   - From \(\frac{\partial \mathcal{L}}{\partial y} = 0\):  
     \[
     -2 = 2\lambda y \Rightarrow \lambda = -\frac{1}{y}
     \]
   - Setting the two expressions for \(\lambda\) equal to each other:
     \[
     \frac{1}{2x} = -\frac{1}{y} \Rightarrow y = -2x
     \]
   - Substitute \(y = -2x\) into the constraint \(x^2 + y^2 = 20\):
     \[
     x^2 + (-2x)^2 = 20 \Rightarrow x^2 + 4x^2 = 20 \Rightarrow 5x^2 = 20 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2
     \]
   - For \(x = 2\), \(y = -4\).  
     For \(x = -2\), \(y = 4\).

4. **Calculate \(f(x, y)\) at the Points**:
   - For \((x, y) = (2, -4)\):  
     \[
     f(2, -4) = 2 - 2(-4) = 2 + 8 = 10
     \]
   - For \((x, y) = (-2, 4)\):  
     \[
     f(-2, 4) = -2 - 2(4) = -2 - 8 = -10
     \]

5. **Identify the Maximum and Minimum Values**:
   - The maximum value is \(10\).
   - The minimum value is \(-10\).

### Final Answers:
- Maximum: \( \mathbf{10} \)
- Minimum: \( \mathbf{-10} \)

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

Here are five additional questions related to this problem:

1. How do Lagrange multipliers help in finding extreme values with constraints?
2. What other methods could you use to solve constrained optimization problems?
3. Why do we set \(\nabla f = \lambda \nabla g\) in the method of Lagrange multipliers?
4. Can we have multiple values for \(\lambda\) in a solution?
5. What is the significance of the constraint \(x^2 + y^2 = 20\)?

**Tip:** When solving constrained optimization problems, always verify that the points satisfy both the constraint and the original function to ensure correct extreme values.

Find the extreme values of f(x, y) = x - 2y subject to g(x, y) = x^2 + y^2 = 20.

Learn how to find the maximum and minimum values of the function f(x, y) = x - 2y subject to the constraint x^2 + y^2 = 20 using the method of Lagrange multipliers. This problem is ideal for students studying multivariable calculus and constrained optimization.

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