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To find the extreme values of the function \( f(x, y) = 2x^2 + 3y^2 - 4x - 7 \) on the region described by \( x^2 + y^2 \leq 16 \), we need to look for both the interior critical points and the boundary points.

### Step 1: Find Interior Critical Points
1. **Compute the partial derivatives** of \( f(x, y) \):
   \[
   f_x = \frac{\partial}{\partial x}(2x^2 + 3y^2 - 4x - 7) = 4x - 4
   \]
   \[
   f_y = \frac{\partial}{\partial y}(2x^2 + 3y^2 - 4x - 7) = 6y
   \]

2. **Set the partial derivatives equal to zero** to find critical points:
   - From \( f_x = 0 \):
     \[
     4x - 4 = 0 \Rightarrow x = 1
     \]
   - From \( f_y = 0 \):
     \[
     6y = 0 \Rightarrow y = 0
     \]

   So, the only critical point inside the region is \( (1, 0) \).

3. **Evaluate \( f \) at the interior critical point**:
   \[
   f(1, 0) = 2(1)^2 + 3(0)^2 - 4(1) - 7 = 2 - 4 - 7 = -9
   \]

### Step 2: Use the Boundary Condition
The boundary of the region is given by \( x^2 + y^2 = 16 \), which represents a circle with radius 4. To analyze the boundary, we can use the method of Lagrange multipliers.

1. **Set up the Lagrange multiplier equations** with \( g(x, y) = x^2 + y^2 - 16 = 0 \):
   \[
   \nabla f = \lambda \nabla g
   \]
   - \( \nabla f = (4x - 4, 6y) \)
   - \( \nabla g = (2x, 2y) \)

   So,
   \[
   4x - 4 = \lambda \cdot 2x
   \]
   \[
   6y = \lambda \cdot 2y
   \]

2. **Solve for \( \lambda \) and the boundary points**:
   - If \( x \neq 0 \):
     \[
     \lambda = \frac{4x - 4}{2x} = 2 - \frac{2}{x}
     \]
   - If \( y \neq 0 \):
     \[
     \lambda = 3
     \]

3. **Evaluate \( f \)

Find the extreme values of f on the region described by the inequality: f(x, y) = 2x^2 + 3y^2 - 4x - 7, x^2 + y^2 ≤ 16.

Learn how to find the maximum and minimum values of the function f(x, y) = 2x^2 + 3y^2 - 4x - 7 on the region defined by x^2 + y^2 ≤ 16. This problem involves finding critical points and using the method of Lagrange multipliers to evaluate boundary conditions. Suitable for undergraduate calculus students.

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