https://file.aiquickdraw.com/mathbot/file/1729457101138y5r956dr.jpg

To solve this problem and find the critical points of the function \( f(x, y) = 2x^2 - 4xy + 5y^2 - 18y \), we will follow these steps:

### Step 1: Find the partial derivatives
We need to compute the first partial derivatives of \( f(x, y) \) with respect to \( x \) and \( y \), and then set them to zero to find the critical points.

- \( f_x(x, y) = \frac{\partial}{\partial x}(2x^2 - 4xy + 5y^2 - 18y) \)
- \( f_y(x, y) = \frac{\partial}{\partial y}(2x^2 - 4xy + 5y^2 - 18y) \)

### Step 2: Solve the system of equations
Once the partial derivatives are set to zero, we will solve the resulting system of equations to get the critical points.

### Step 3: Classify the critical points
To classify the critical points, we will use the second partial derivatives and the **Hessian matrix** to calculate the discriminant \( D = f_{xx}f_{yy} - (f_{xy})^2 \). Depending on the value of \( D \), we can classify the points as local minima, local maxima, or saddle points.

Let me compute this step-by-step for you.

### Step 1: Partial Derivatives
1. \( f_x(x, y) = \frac{\partial}{\partial x}(2x^2 - 4xy + 5y^2 - 18y) = 4x - 4y \)
2. \( f_y(x, y) = \frac{\partial}{\partial y}(2x^2 - 4xy + 5y^2 - 18y) = -4x + 10y - 18 \)

### Step 2: Solve for Critical Points
Set \( f_x(x, y) = 0 \) and \( f_y(x, y) = 0 \) to find the critical points:
- \( 4x - 4y = 0 \)  →  \( x = y \)
- \( -4x + 10y - 18 = 0 \)

Substitute \( x = y \) into the second equation:
- \( -4y + 10y - 18 = 0 \)
- \( 6y - 18 = 0 \)  →  \( y = 3 \)

So, \( y = 3 \), and since \( x = y \), \( x = 3 \).

Thus, the critical point is \( (3, 3) \).

### Step 3: Classify the Critical Point
We now compute the second partial derivatives to classify the critical point:
- \( f_{xx} = \frac{\partial}{\partial x}(4x - 4y) = 4 \)
- \( f_{yy} = \frac{\partial}{\partial y}(-4x + 10y - 18) = 10 \)
- \( f_{xy} = \frac{\partial}{\partial y}(4x - 4y) = -4 \)

Now, calculate the discriminant:
\[
D = f_{xx}f_{yy} - (f_{xy})^2 = (4)(10) - (-4)^2 = 40 - 16 = 24
\]
Since \( D > 0 \) and \( f_{xx} > 0 \), the critical point \( (3, 3) \) is a **local minimum**.

### Conclusion:
- **Number of critical points**: 1
- **Classification**: Local minimum at \( (3, 3) \)

Would you like more details or have any questions?

Here are 5 related questions you might find helpful:
1. How do you determine if a critical point is a local maximum or minimum using the Hessian matrix?
2. What is the significance of the discriminant \( D \) in classifying critical points?
3. Could the function have saddle points or maxima if the conditions were different?
4. How do we interpret the results geometrically in terms of the surface \( f(x, y) \)?
5. How would adding higher-order terms in the function affect the classification of critical points?

**Tip**: Always check both the sign of \( D \) and the second partial derivative with respect to \( x \) (i.e., \( f_{xx} \)) when classifying critical points.

Find the critical point(s) and classify them as local maxima, local minima, or saddle points for the function f(x, y) = 2x^2 - 4xy + 5y^2 - 18y.

Learn how to find and classify the critical points of the multivariable function f(x, y) = 2x^2 - 4xy + 5y^2 - 18y. Using partial derivatives and the Hessian matrix, this calculus problem explores local minima, maxima, and saddle points. Suitable for university-level calculus students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation