How do I find critical points and determine local extrema for a given function?

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Homework Help Overview

The discussion revolves around finding critical points and determining local extrema for the function f(x,y)=4xy−(x^4)/2−y^2. Participants are exploring the process of calculating partial derivatives and applying the second derivative test to classify critical points.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss setting partial derivatives to zero to find critical points, leading to the identification of points (0,0), (-2,-4), and (2,4). There is an exploration of the second derivative test to classify these points, with some questioning the absence of local minima.

Discussion Status

The discussion is active, with participants sharing their calculations and findings. There is acknowledgment of a potential misunderstanding regarding the classification of critical points, particularly concerning the inconclusive result at (0,0). Some participants are clarifying their calculations and assumptions.

Contextual Notes

Participants are working under the constraints of homework guidelines, which may limit the depth of exploration. There is a noted confusion about the role of the mixed partial derivative in the classification process.

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Homework Statement


find all critical points and identify the locations of local maximums, minimums and saddle points of the function f(x,y)=4xy-\frac{x^4}{2}-y^2

Homework Equations

The Attempt at a Solution


setting Partial derivative respect to x = 0 : 4y-\frac{4x^3}{2}=0
setting partial derivative respect to y=0: 4x-2y=0

from the second equation, Y=2X, plug it into the first equation. I get : 2x(4-x^2)=0
now, x=0 or x=-2 or x=2. since y=2x, now I got 3 sets of critical points. (0,0), (-2,-4) and (2,4)

second partial derivative respect to x = -6x^2
second partial derivative respect to y=-2
and ƒxy=0

so D(a,b)= 12x^2

plug in these critical points into D(a,b) I found that local max value at (-2,-4) and (2,4) for which both give D value of 48, greater than 0, and ƒxx at both points are smaller than 0.
for D(0,0) , the test is inconclusive .

I did not find any local min, I am wondering if i missed something.
any help is appreciated
 
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qq545282501 said:

Homework Statement


find all critical points and identify the locations of local maximums, minimums and saddle points of the function f(x,y)=4xy-\frac{x^4}{2}-y^2

second partial derivative respect to x = -6x^2
second partial derivative respect to y=-2
and ƒxy=0

so D(a,b)= 12x^2
fxy is not zero.
 
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Likes   Reactions: qq545282501
qq545282501 said:

Homework Statement


find all critical points and identify the locations of local maximums, minimums and saddle points of the function f(x,y)=4xy-\frac{x^4}{2}-y^2

Homework Equations

The Attempt at a Solution


setting Partial derivative respect to x = 0 : 4y-\frac{4x^3}{2}=0
setting partial derivative respect to y=0: 4x-2y=0

from the second equation, Y=2X, plug it into the first equation. I get : 2x(4-x^2)=0
now, x=0 or x=-2 or x=2. since y=2x, now I got 3 sets of critical points. (0,0), (-2,-4) and (2,4)

second partial derivative respect to x = -6x^2
second partial derivative respect to y=-2
and ƒxy=0

so D(a,b)= 12x^2

plug in these critical points into D(a,b) I found that local max value at (-2,-4) and (2,4) for which both give D value of 48, greater than 0, and ƒxx at both points are smaller than 0.
for D(0,0) , the test is inconclusive .

I did not find any local min, I am wondering if i missed something.
any help is appreciated

##f_{xy} \neq 0##.
 
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Likes   Reactions: qq545282501
opps, got it. thank you
 

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