Math Problem Statement

Find and classify the stationary values of x^2+y^2-4x+6y+12=0

Solution

To find and classify the stationary values of the function f(x,y)=x2+y24x+6y+12f(x, y) = x^2 + y^2 - 4x + 6y + 12, we will proceed through the following steps:

1. Formulate the function:

We start with the function: f(x,y)=x2+y24x+6y+12f(x, y) = x^2 + y^2 - 4x + 6y + 12

2. Find the first partial derivatives:

We calculate the partial derivatives of f(x,y)f(x, y) with respect to xx and yy:

fx=2x4\frac{\partial f}{\partial x} = 2x - 4 fy=2y+6\frac{\partial f}{\partial y} = 2y + 6

3. Set the partial derivatives equal to zero:

To find the stationary points, we set these partial derivatives equal to zero:

2x4=0x=22x - 4 = 0 \quad \Rightarrow \quad x = 2 2y+6=0y=32y + 6 = 0 \quad \Rightarrow \quad y = -3

Thus, the stationary point is (x,y)=(2,3)(x, y) = (2, -3).

4. Second partial derivatives and Hessian matrix:

To classify the stationary point, we compute the second partial derivatives:

2fx2=2,2fy2=2,2fxy=0\frac{\partial^2 f}{\partial x^2} = 2, \quad \frac{\partial^2 f}{\partial y^2} = 2, \quad \frac{\partial^2 f}{\partial x \partial y} = 0

Now, we construct the Hessian matrix HH:

\frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \\ \frac{\partial^2 f}{\partial x \partial y} & \frac{\partial^2 f}{\partial y^2} \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}$$ ### 5. **Classify the stationary point:** To classify the stationary point, we evaluate the determinant of the Hessian matrix: $$\det(H) = (2)(2) - (0)(0) = 4$$ Since the determinant of the Hessian matrix is positive and $$\frac{\partial^2 f}{\partial x^2} = 2 > 0$$, the stationary point $$(2, -3)$$ is a **local minimum**. ### Conclusion: The function has a stationary point at $$(2, -3)$$, and this point is classified as a **local minimum**. Would you like more details or have any further questions? Here are 5 related questions you might find interesting: 1. How do you classify stationary points using the Hessian matrix? 2. What happens if the Hessian determinant is zero? 3. Can stationary points be saddle points, and how do we detect them? 4. How do we handle stationary points in multivariable optimization problems? 5. What is the significance of eigenvalues of the Hessian matrix? **Tip:** Always check both the determinant and the second derivative signs when classifying stationary points!

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Math Problem Analysis

Mathematical Concepts

Multivariable Calculus
Stationary Points
Optimization

Formulas

f(x, y) = x^2 + y^2 - 4x + 6y + 12
Partial derivatives: ∂f/∂x = 2x - 4, ∂f/∂y = 2y + 6
Hessian matrix: H = [[2, 0], [0, 2]]
Determinant of Hessian: det(H) = 4

Theorems

Stationary points classification using second derivative test
Hessian matrix determinant test

Suitable Grade Level

Undergraduate Calculus (Grades 11-12 or early college)