Suppose that f(x, y) = 4x² + 4y¹ then the minimum is -2xy
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
![**Identifying Critical Points Using the Second Derivative**
To determine whether a critical point of a function is a local maximum, local minimum, or saddle point, the second derivative test is employed.
### Problem Statement:
Suppose the function is given by:
\[ f(x, y) = 4x^4 + 4y^4 - 2xy \]
Determine the minimum value of \( f(x, y) \).
### Answer:
- The minimum value of the function is located in the blank space provided.
### Additional Resources:
- **Video**: A video resource is available for further explanation.
- **Message Instructor**: If you need more help, you can message the instructor for guidance.
By analyzing the function's second derivatives, particularly the Hessian matrix, and evaluating it at critical points, one can classify these points as local maximum, local minimum, or saddle points.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb8216494-bf6c-44b6-949f-dd8ff9f3594a%2F20bc4e56-206a-4d28-a339-1ac36c934db9%2Folgidrv_processed.png&w=3840&q=75)
Transcribed Image Text:**Identifying Critical Points Using the Second Derivative**
To determine whether a critical point of a function is a local maximum, local minimum, or saddle point, the second derivative test is employed.
### Problem Statement:
Suppose the function is given by:
\[ f(x, y) = 4x^4 + 4y^4 - 2xy \]
Determine the minimum value of \( f(x, y) \).
### Answer:
- The minimum value of the function is located in the blank space provided.
### Additional Resources:
- **Video**: A video resource is available for further explanation.
- **Message Instructor**: If you need more help, you can message the instructor for guidance.
By analyzing the function's second derivatives, particularly the Hessian matrix, and evaluating it at critical points, one can classify these points as local maximum, local minimum, or saddle points.
Expert Solution
Step 1
1st we will find critical points
At critical point
fx ( partial differentiation with respect to x ) = 0
and fy ( partial differentiation with respect to y ) = 0
Find values of (x,y)
Then calculate D = fxx .fyy - (fxy)2 at critical points
If D>0 and fxx > 0 => relative minimum
If D>0 and fxx < 0 => relative maximum
If D<0 => saddle point
If D = 0 then no conclusion can be drawn
Step by step
Solved in 4 steps
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