The correct answer is: A. is a minimum equal to $\frac{{10}}{3}$.
To find the optimal value of a function, we can use the method of Lagrange multipliers. The Lagrange function for this problem is:
$$L(x, y, \lambda) = 4x^2 + 6y^2 – 8x – 4y + 8 + \lambda(4x – 8y)$$
Setting the partial derivatives of $L$ with respect to $x$, $y$, and $\lambda$ equal to zero, we get the system of equations:
$$\begin{align}
8x – 8\lambda &= 0 \
6y – 8\lambda &= 0 \
4x – 8y &= 0
\end{align}$$
Solving this system of equations, we get the critical points $(x, y) = \left(\frac{1}{2}, \frac{1}{3}\right)$.
To evaluate the function $f$ at the critical point, we get:
$$f\left(\frac{1}{2}, \frac{1}{3}\right) = \frac{{10}}{3}$$
Since the critical point is a minimum, we know that the optimal value of $f$ is $\frac{{10}}{3}$.
Here is a brief explanation of each option:
- Option A: The optimal value of $f(x, y)$ is a minimum equal to $\frac{{10}}{3}$. This is correct, as shown above.
- Option B: The optimal value of $f(x, y)$ is a maximum equal to $\frac{{10}}{3}$. This is incorrect, as the optimal value is a minimum.
- Option C: The optimal value of $f(x, y)$ is a minimum equal to $\frac{8}{3}$. This is incorrect, as the optimal value is $\frac{{10}}{3}$.
- Option D: The optimal value of $f(x, y)$ is a maximum equal to $\frac{8}{3}$. This is incorrect, as the optimal value is a minimum.