Find the local extrema of f(x,y)=-2y^2+17y-2x^2-13x-4

To find the local extrema of the function \( f(x, y) = -2y^2 + 17y - 2x^2 - 13x - 4 \), we'll follow these steps:

### 1. **Compute the partial derivatives:**
   - First, we find the partial derivative of \( f(x, y) \) with respect to \( x \):
     \[
     f_x = \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \left( -2x^2 - 13x - 2y^2 + 17y - 4 \right) = -4x - 13
     \]
   - Next, we find the partial derivative of \( f(x, y) \) with respect to \( y \):
     \[
     f_y = \frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \left( -2y^2 + 17y - 2x^2 - 13x - 4 \right) = -4y + 17
     \]

### 2. **Set the partial derivatives equal to zero:**
   - Set \( f_x = 0 \) and \( f_y = 0 \) to find the critical points:
     \[
     -4x - 13 = 0
     \]
     \[
     -4y + 17 = 0
     \]
   
   - Solving these equations:
     \[
     x = -\frac{13}{4} = -3.25
     \]
     \[
     y = \frac{17}{4} = 4.25
     \]
     
   So, the critical point is \( \left( x, y \right) = \left( -3.25, 4.25 \right) \).

### 3. **Determine the nature of the critical point:**
   - To classify the critical point, we compute the second partial derivatives and use the second derivative test.
     - Second partial derivative with respect to \( x \):
       \[
       f_{xx} = \frac{\partial^2 f}{\partial x^2} = -4
       \]
     - Second partial derivative with respect to \( y \):
       \[
       f_{yy} = \frac{\partial^2 f}{\partial y^2} = -4
       \]
     - Mixed partial derivative:
       \[
       f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 0
       \]
   
   - The discriminant \( D \) is given by:
     \[
     D = f_{xx}f_{yy} - (f_{xy})^2 = (-4)(-4) - (0)^2 = 16
     \]
     
   - Since \( D > 0 \) and \( f_{xx} < 0 \), the function has a **local maximum** at the critical point.

### 4. **Conclusion:**
   The local maximum of the function \( f(x, y) \) occurs at \( \left( x, y \right) = \left( -3.25, 4.25 \right) \).

Would you like more details on this solution, or do you have any questions?

### Relative Questions:
1. What are the conditions under which a critical point is a saddle point?
2. How do you find the global extrema for a function?
3. What is the significance of the discriminant \( D \) in classifying critical points?
4. How would the solution change if \( f(x, y) \) had linear terms only?
5. Can there be more than one local maximum in a multivariable function?
6. How can you extend this analysis to functions of three or more variables?
7. What if the second derivative test is inconclusive? How should you proceed?
8. How would you find the extrema of a function subject to a constraint?

### Tip:
When using the second derivative test, always check the sign of the discriminant \( D \) and the second partial derivatives to classify the critical points correctly.

Discover how to find the local extrema of the function f(x, y) = -2y^2 + 17y - 2x^2 - 13x - 4 using partial derivatives and the second derivative test. This problem involves critical points, second partial derivatives, and discriminants. Suitable for advanced undergraduate students and anyone interested in multivariable calculus.

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