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GDB No.2
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Question:
Locate all the extrema and saddle points of
f(x,y)=3xy-x³-y³
Kindly Discuss here. Jazak Allah.
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Very easy ...... !!!
1) First u will find 1st order partial derivative.
2 )then 2nd order partial derivative.
3) Find the critical points setting 1st order partial derivative = 0
4) Then Solve the equation that u get from above step and find the coordinate points .
5) U will get 2 points . Now put these points in 2nd order partial derivative.
6) Now put these values in D (,) = ...... and check your pints for saddle and relative max or relative min .....
Simple ........
Have done and Submitted ..
File Me upload ai Kar Do!
Basically MTH301 is not mine Subject .... U should do your work yourself . Where u will stuck u can ask for help .. I will definitely help u ..
Hints ... U will get two points .. (0,0) and (1,1)
(0,0) is saddle point and (1,1) is relative maximum ...
question ye ha mere pas idhar wrong diya hua baqi sab logo k pas yehi ha kiya
f(x,y)=3xy−x^3−y^3
yup thats the solution
n D can be written as
D(x,y)=fxx . fyy - f^2 xy
if value will ve negative then its saddle point,
if positive n 2nd order derivative is negative it relative maximum
if d is positive n 2nd derivative is positive its relative minimum
Sorry for a little mistake...
The right question is:
f(x,y)=3xy−x^3−y^3
Please Discuss here about this GDB.Thanks
Our main purpose here discussion not just Solution
We are here with you hands in hands to facilitate your learning and do not appreciate the idea of copying or replicating solutions.
mth301 gdb 2 solution here
Misbha thanks for sharing
kindly ye file ider copy paste kr k post kr de koi mere laptop mein file open ni ho rahi
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