Question

Solve please ?

Answer 1

`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`