Math Problem Statement

please use dy/dx as formula

Find the critical points and determine whether minimum or maximum for the function: f(x, y) = 2x^2 + 3y^2 + 9y + 2x + 5xy

Solution

The problem requires finding the critical points of the given function f(x,y)=2x2+3y2+9y+2x+5xyf(x, y) = 2x^2 + 3y^2 + 9y + 2x + 5xy, and determining whether the critical points correspond to a local minimum, maximum, or saddle point. Let's proceed step by step:


Step 1: Compute Partial Derivatives

To find the critical points, we need the partial derivatives of f(x,y)f(x, y) with respect to xx and yy:

fx=4x+5y+2\frac{\partial f}{\partial x} = 4x + 5y + 2 fy=6y+5x+9\frac{\partial f}{\partial y} = 6y + 5x + 9


Step 2: Solve for Critical Points

Set the partial derivatives to zero:

  1. fx=0\frac{\partial f}{\partial x} = 0:
    4x+5y+2=04x + 5y + 2 = 0

  2. fy=0\frac{\partial f}{\partial y} = 0:
    6y+5x+9=06y + 5x + 9 = 0

We now solve this system of equations for xx and yy. Substituting xx or yy from one equation into the other will provide the critical points.


Step 3: Analyze Critical Points

Use the second derivative test to determine the nature of the critical points. First, compute the second partial derivatives:

  • fxx=2fx2=4f_{xx} = \frac{\partial^2 f}{\partial x^2} = 4
  • fyy=2fy2=6f_{yy} = \frac{\partial^2 f}{\partial y^2} = 6
  • fxy=fyx=2fxy=5f_{xy} = f_{yx} = \frac{\partial^2 f}{\partial x \partial y} = 5

The Hessian determinant is given by: H=fxxfyy(fxy)2=(4)(6)(5)2=2425=1H = f_{xx} f_{yy} - (f_{xy})^2 = (4)(6) - (5)^2 = 24 - 25 = -1

Since H<0H < 0, the critical point (once solved) corresponds to a saddle point.

Let me calculate the critical points explicitly. sympy import symbols, Eq, solve

Define variables

x, y = symbols('x y')

Partial derivatives set to zero

partial_x = 4x + 5y + 2 partial_y = 6y + 5x + 9

Solve the system of equations

critical_points = solve([Eq(partial_x, 0), Eq(partial_y, 0)], (x, y)) critical_pointsThe critical point is (33,26)(-33, 26).

Step 4: Nature of the Critical Point

As previously calculated, the Hessian determinant (H=1H = -1) is negative, which means the critical point (33,26)(-33, 26) is a saddle point.


Would you like a detailed explanation of any specific step? Here are some related questions:

  1. How do you compute second partial derivatives for multivariable functions?
  2. What is the Hessian determinant, and how does it determine the nature of a critical point?
  3. Why is a saddle point neither a maximum nor a minimum?
  4. Can this method be extended to higher dimensions (e.g., functions of three variables)?
  5. How would graphing this function confirm the nature of the critical point?

Tip: Always verify the determinant of the Hessian to confirm your interpretation of critical points, especially when analyzing complex functions.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Critical Points
Partial Derivatives
Second Derivative Test
Hessian Matrix

Formulas

Partial derivative: ∂f/∂x and ∂f/∂y
Second derivative test: Hessian determinant H = f_xx * f_yy - (f_xy)^2

Theorems

Second Derivative Test

Suitable Grade Level

Undergraduate (Calculus Level)