The problem is asking to find the extreme values of the function:

$f(x, y) = 2x^2 + 3y^2 - 4x - 8$

on the region described by the inequality:

$x^2 + y^2 \leq 16$

This describes a closed disk with radius 4 and center at the origin. We need to find the maximum and minimum values of $f(x, y)$ inside this region.

Steps to Solve:

1. Find the Critical Points:

   • Compute the partial derivatives $f_x$ and $f_y$ and solve $f_x = 0$ and $f_y = 0$ to find critical points inside the region.

2. Analyze the Boundary:

   • Use the method of Lagrange multipliers to find extrema on the boundary of the disk, which is given by $x^2 + y^2 = 16$.

3. Compare Values:

   • Evaluate $f(x, y)$ at the critical points and on the boundary, and determine the maximum and minimum values.

Let's start with the first step by computing the partial derivatives of $f(x, y)$.

The function is:

$f(x, y) = 2x^2 + 3y^2 - 4x - 8$

Partial Derivatives:

• $f_x = \frac{\partial}{\partial x}(2x^2 + 3y^2 - 4x - 8) = 4x - 4$
• $f_y = \frac{\partial}{\partial y}(2x^2 + 3y^2 - 4x - 8) = 6y$

Now, let's find the critical points by setting these partial derivatives equal to zero:

$4x - 4 = 0 \quad \text{and} \quad 6y = 0$

Solving this system:

$x = 1, \quad y = 0$

Thus, the critical point is $(1, 0)$.

Check if the critical point is inside the region:

The region is $x^2 + y^2 \leq 16$. At $(1, 0)$:

$1^2 + 0^2 = 1 \leq 16$

So, $(1, 0)$ lies inside the region.

Evaluate $f(x, y)$ at the critical point:

$f(1, 0) = 2(1)^2 + 3(0)^2 - 4(1) - 8 = 2 - 4 - 8 = -10$

Now let's move to the boundary of the region using Lagrange multipliers. We will analyze that next.The solutions to the Lagrange multiplier system provide the points on the boundary of the disk where potential extrema might occur. These points are:

1. $(-4, 0)$
2. $(-2, -2\sqrt{3})$
3. $(-2, 2\sqrt{3})$
4. $(4, 0)$

Now, we will evaluate the function $f(x, y)$ at these points to compare their values.

Additionally, we already found the critical point inside the region at $(1, 0)$. Let's compute the values of $f(x, y)$ at all these points.The values of $f(x, y)$ at the critical point and the boundary points are:

1. $f(1, 0) = -10$ (Critical point inside the region)
2. $f(-4, 0) = 40$
3. $f(-2, -2\sqrt{3}) = 44$
4. $f(-2, 2\sqrt{3}) = 44$
5. $f(4, 0) = 8$

Conclusion:

• The minimum value of $f(x, y)$ is $-10$ at the point $(1, 0)$.
• The maximum value of $f(x, y)$ is $44$ at the points $(-2, -2\sqrt{3})$ and $(-2, 2\sqrt{3})$.

Would you like further clarification on any of the steps?

Here are some related questions to consider:

1. How would the solution change if the region were a different shape?
2. How do Lagrange multipliers work in higher dimensions?
3. Can the method of Lagrange multipliers be applied if the constraint is an inequality instead of an equality?
4. How would we approach finding extrema if the boundary were more complex, like a non-circular shape?
5. What is the geometric interpretation of the critical point and boundary points in this problem?

Tip: When solving optimization problems with constraints, always verify if the critical points lie within the feasible region before proceeding to boundary analysis.