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December 20, 2014

December 20, 2014

Posted by **Lucy** on Monday, February 27, 2012 at 12:54am.

- cal3 -
**Steve**, Monday, February 27, 2012 at 12:38pmF = -3yx^2-3xy^2+36xy

Fx = -6xy - 3y^2 + 36y

Fxx = -6y

Fy = -3x^2 - 6xy + 36x

Fyy = -6x

Fxy = -6x - 6y + 36

D = FxxFyy-(Fxy)^2 = 36xy - 36(x+y-6)^2

Fx = 0 Fy=0 at (0,0)

D<0 so a saddle point

Fx = -6xy - 3y^2 + 36y

Fx = 0 when y = 2(6-x)

Fy = -3x^2 - 6xy + 36x

Fy = 0 when x = 2(6-y)

So there is a local max for z along those two lines

- cal3 -
**Lucy**, Monday, February 27, 2012 at 6:12pmOkay Im just confused as how to get Fx=0 and Fy=0, If I set -6xy-3y^2+36y=0 how do i solve this?? Algebra was so long ago!

could you detail that part of the problem...

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