Maxima and minima of functions of two variables

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Homework Help Overview

The discussion revolves around identifying relative maxima, minima, and saddle points for the function f(x,y) = ysin(x). Participants are analyzing critical points and the second derivative test for functions of two variables.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the critical points derived from the first derivatives and question the implications of the second derivatives. There is confusion regarding the calculation of fyy and the interpretation of the results, particularly concerning the presence of saddle points.

Discussion Status

The conversation is ongoing, with participants clarifying their understanding of the derivatives and the conditions for identifying saddle points. Some express uncertainty about the calculations and the implications of setting y to zero in the function.

Contextual Notes

There appears to be a misunderstanding regarding the second derivative test and the necessary conditions for determining relative extrema and saddle points. Participants are also addressing potential typos in their calculations.

Mdhiggenz
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Homework Statement



Locate all relative maxima, relative minima ,and saddle points if any.

f(x,y)=ysinx

fx(x,y)=ycosx

fy(x,y)=sinx

ycosx=0 sinx=0
y=0 x=0,∏,2∏... up until infinity
Critical points at (0,0),(∏,0),(2∏,0)...

fxx(x,y)=-ysinx

fyy(x,y)= cosx

fxx*fyy-f(x,y)2→0-0=0 ∴ no relative extrema,

however the book says that fxx*fyy-f(x,y)2=-1 which means saddle point.

I don't understand if y is = to zero.. the function ysinx will always be zero..

Homework Equations





The Attempt at a Solution

 
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Fyy=0
 
Sorry that was a typo, but still doesn't help regardless.
 
Mdhiggenz said:

Homework Statement



Locate all relative maxima, relative minima ,and saddle points if any.

f(x,y)=ysinx

fx(x,y)=ycosx

fy(x,y)=sinx

ycosx=0 sinx=0
y=0 x=0,∏,2∏... up until infinity
Critical points at (0,0),(∏,0),(2∏,0)...

fxx(x,y)=-ysinx

fyy(x,y)= cosx

fxx*fyy-f(x,y)2→0-0=0 ∴ no relative extrema,

however the book says that fxx*fyy-f(x,y)2=-1 which means saddle point.

I don't understand if y is = to zero.. the function ysinx will always be zero..

Homework Equations



The Attempt at a Solution

Isn't fy(x,y) = sin(x) ?

Then what is fyy(x,y) ? Isn't it zero?

Also, What is fxy(x,y) ?

You need fxx*fyy-(fxy)2 , not fxx*fyy-f(x,y)2
 
Last edited:

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