Math Problem Statement
Find the local maximum and minimum values, and saddle point(s) of f(x,y). f(x,y)=y^3-3y+3x^2y.
Solution
To find the local maximum, local minimum, and saddle points of the function , we follow these steps:
1. Find the Critical Points
First, we need to find the partial derivatives of and set them to zero.
Partial Derivatives:
Set these partial derivatives equal to zero to find the critical points.
Solve :
This gives two possibilities:
Solve :
Combine the conditions:
-
For : So, the critical points are and .
-
For : So, the critical points are and .
2. Classify the Critical Points
We use the second derivative test, which involves computing the Hessian matrix.
Second Partial Derivatives:
The Hessian matrix is:
f_{xx} & f_{xy} \\ f_{xy} & f_{yy} \end{pmatrix} = \begin{pmatrix} 6y & 6x \\ 6x & 6y \end{pmatrix}$$ The determinant of $$H$$ is: $$\text{det}(H) = (6y)(6y) - (6x)(6x) = 36y^2 - 36x^2 = 36(y^2 - x^2)$$ Evaluate the determinant at each critical point: 1. At $$(0, 1)$$: $$\text{det}(H) = 36(1^2 - 0^2) = 36 > 0$$ $$f_{xx} = 6 \cdot 1 = 6 > 0$$ So, $$(0, 1)$$ is a local minimum. 2. At $$(0, -1)$$: $$\text{det}(H) = 36((-1)^2 - 0^2) = 36 > 0$$ $$f_{xx} = 6 \cdot (-1) = -6 < 0$$ So, $$(0, -1)$$ is a local maximum. 3. At $$(1, 0)$$ and $$(-1, 0)$$: $$\text{det}(H) = 36(0^2 - 1^2) = -36 < 0$$ So, $$(1, 0)$$ and $$(-1, 0)$$ are saddle points. ### Summary - **Local Maximum**: $$(0, -1)$$ - **Local Minimum**: $$(0, 1)$$ - **Saddle Points**: $$(1, 0)$$ and $$(-1, 0)$$ Would you like more details or have any questions? Here are some related questions you might be interested in: 1. How do you find critical points for functions of multiple variables? 2. What is the Hessian matrix and how is it used? 3. Can you show examples of saddle points in different functions? 4. How do you determine if a critical point is a local minimum, maximum, or saddle point? 5. What are the applications of finding local maxima and minima in real-world problems? **Tip:** When analyzing functions of several variables, always check the second partial derivatives and the determinant of the Hessian matrix to classify the critical points accurately.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Multivariable Calculus
Critical Points
Second Derivative Test
Formulas
Partial derivatives
Hessian matrix
Determinant of Hessian
Theorems
Second Derivative Test
Suitable Grade Level
Advanced Undergraduate
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