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 $f(x, y) = 8x^2 + y^2$: