critical points exist where fx & fy=0
fx=-36-1/(x+y) =0 ... 1/(x+y)=-36
fy=-12y -1/(x+y) =0 ... y= 3 and x= -109/36
fxx=0 fyy=-12 fxy=0 thus D=0*-12-02=0 ... indeterminate
Virginia G.
asked 04/03/23Find the critical point of the function f(x,y)=−(36x+6y^2+ln(|x+y|)) .
Use the Second Derivative Test to determine whether it is a local min, max, saddle point or test fails
critical points exist where fx & fy=0
fx=-36-1/(x+y) =0 ... 1/(x+y)=-36
fy=-12y -1/(x+y) =0 ... y= 3 and x= -109/36
fxx=0 fyy=-12 fxy=0 thus D=0*-12-02=0 ... indeterminate
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