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

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The discussion focuses on finding critical points and local extrema for the function f(x,y) = 4xy - (x^4/2) - y^2. The critical points identified are (0,0), (-2,-4), and (2,4) after setting the partial derivatives to zero. The second derivative test reveals that (-2,-4) and (2,4) are local maxima, while the point (0,0) yields an inconclusive result. The participant expresses uncertainty about the absence of local minima and seeks clarification. The analysis confirms that the function has local maxima but no local minima.
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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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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 qq545282501
opps, got it. thank you
 
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