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If f(x,y)=2y^2x-yx^2+4xy  , find the local extrema and saddle points of  f .

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owlstalk | Student, Undergraduate | eNoter

Posted November 24, 2010 at 2:16 PM via web

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If f(x,y)=2y^2x-yx^2+4xy  , find the local extrema and saddle points of  f .

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kjcdb8er | Teacher | (Level 1) Associate Educator

Posted November 29, 2010 at 3:39 AM (Answer #1)

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higher dimention problems are solved analogously with 2-D extrema problems: take derivatives for each direction, and solve.

Steps:

Find critical points where Fx = Fy = 0; evaluate second derivative at those points:

. Fxx < 0  & Fxx Fyy -  Fxy^2 > 0 --> maximum

. Fxx < 0 & Fxx Fyy - Fxy^2 > 0  --> minimum

. Fxx Fyy - Fxy^2 < 0  --> saddle point

 

f(x,y) = 2y^2x - yx^2 + 4xy

fx = 2y^2 - 2yx + 4y    -->  fxx = -2y  and  fxy = 4y - 2x + 4

fy = 4yx - x^2 + 4x   --> fyy = 4y   and  fyx = 4y - 2x + 4

Note that fxy = fyx, as it should.

 

Solve system for critical points:

fx = 2y^2 - 2yx + 4y = 0

fy = 4yx - x^2 + 4x = 0

solutions: (0,-2), (4/3, -2/3), (4,0), (0,0)

 

We'll complete the solution for one of these points. Each of the points is evaluated in like manner:

fxx (4/3, -2/3) = -2y = -2(-2/3) = 4/3 > 0

fyy(4/3, -2/3) = 4y = 4(-2/3) = -8/3

fxy(4/3, -2/3) = 4y - 2x + 4 = -8/3 - 8/3 + 4 = -4/3

fxx fyy - fxy^2 = 4/3 * -8/3 - 16/9 = -16/3 < 0

Because fxx fyy - fxy^2 < 0, the point (4/3, -2/3) is a saddle point.

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