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1 Answer
Cesareo R.
Oct 18, 2017

#2#

Explanation:

Given #f(x,y) = x^2+4x y+6y^2-4y+4#

The local stationary points are the solutions for

#grad f(x,y) = (f_x,f_y) = (2x+4y,4x+12y-4) = (0,0)#

then the stationary point's set is #(x,y) =(-2,1)#

The stationary point's qualification is given analyzing the eigenvalues for #grad^2f = H# so

#H = 2((1,2),(2,6))# with characteristic polynomial

#p(lambda)=8 - 14 lambda + lambda^2# and with roots

#lambda = 7 pm sqrt41# both positive, then the stationary point at #(-2,1)# is a local minimum point. It is also a global minimum point and #f(-2,1) = 2#

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