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To solve this problem, we’ll use the method of **Lagrange multipliers** to find the minimum and maximum values of \( f(x, y) = 8x^2 + y^2 \) subject to the constraint \( 4x + 2y = 2 \).

### Step 1: Set Up the Lagrange Function
Define the **Lagrange function** \( \mathcal{L} \) by incorporating the constraint into the function with a multiplier \( \lambda \):
\[
\mathcal{L}(x, y, \lambda) = 8x^2 + y^2 + \lambda (2 - 4x - 2y)
\]

### Step 2: Compute the Partial Derivatives
To find the critical points, take partial derivatives of \( \mathcal{L} \) with respect to \( x \), \( y \), and \( \lambda \), and set them equal to zero:
1. \( \frac{\partial \mathcal{L}}{\partial x} = 16x - 4\lambda = 0 \)
2. \( \frac{\partial \mathcal{L}}{\partial y} = 2y - 2\lambda = 0 \)
3. \( \frac{\partial \mathcal{L}}{\partial \lambda} = 2 - 4x - 2y = 0 \)

### Step 3: Solve the System of Equations
From equations (1) and (2), we get:
1. \( 16x = 4\lambda \Rightarrow \lambda = 4x \)
2. \( 2y = 2\lambda \Rightarrow \lambda = y \)

Now, equate \( \lambda \) values from both equations:
\[
4x = y
\]

Substitute \( y = 4x \) into the constraint \( 4x + 2y = 2 \):
\[
4x + 2(4x) = 2 \Rightarrow 12x = 2 \Rightarrow x = \frac{1}{6}
\]

Then, substitute \( x = \frac{1}{6} \) back to find \( y \):
\[
y = 4 \left(\frac{1}{6}\right) = \frac{2}{3}
\]

### Step 4: Evaluate \( f(x, y) \) at \( \left( \frac{1}{6}, \frac{2}{3} \right) \)
Substitute \( x = \frac{1}{6} \) and \( y = \frac{2}{3} \) into \( f(x, y) = 8x^2 + y^2 \):
\[
f\left(\frac{1}{6}, \frac{2}{3}\right) = 8\left(\frac{1}{6}\right)^2 + \left(\frac{2}{3}\right)^2
= 8 \cdot \frac{1}{36} + \frac{4}{9}
= \frac{2}{9} + \frac{4}{9} = \frac{6}{9} = \frac{2}{3}
\]

### Conclusion
The minimum value of \( f(x, y) \) is \( \frac{2}{3} \). Since there are no other critical points, this is the only extremum, and no maximum exists.

**Answers:**
- \( f_{\text{min}} = \frac{2}{3} \)
- \( f_{\text{max}} = \text{DNE} \)

Would you like further details or have any questions?

Here are some related questions you might find helpful:
1. What are Lagrange multipliers, and why are they useful?
2. How do we determine if an extremum exists when using constraints?
3. What is the geometric interpretation of the constraint \( 4x + 2y = 2 \)?
4. Can this method apply to more than two variables and constraints?
5. How would the process change if we were maximizing \( f(x, y) \)?

**Tip:** When setting up a Lagrange multiplier problem, remember to carefully derive and solve each partial derivative equation systematically.

Find the minimum and maximum values of the function f(x, y) = 8x^2 + y^2, subject to the constraint 4x + 2y = 2.

Discover how to find the minimum and maximum values of f(x, y) = 8x^2 + y^2 subject to the constraint 4x + 2y = 2. This solution uses the Lagrange multipliers method, suitable for college-level calculus students.

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