Question

Find the local extreme values of the function f(x, y) = xy - x2 - y2 - 3x - 3y + 12

Answer 1

`f(x,y)=xy-x^2-y^2-3x-3y+12`

First compute the first-order partial derivatives and find the critical points.



`f_x=y-2x-3`

`f_y=x-2y-3`

Both first order derivatives vanish at
`(x,y)=(-3,-3)`.


Computing the Hessian, we get


`\mathbf H(x,y)=\begin{bmatrix}f_(xx)&f_(xy)\\f_(yx)&f_(yy)\end{bmatrix}=\begin{bmatrix}-2&1\\1&-2\end{bmatrix}`

We have
`\det\mathbf H(x,y)=3>0`, which means
`(-3,-3)` is an extremum of
`f(x,y)`. Since
`f_(xx)(-3,-3)=-2<0`, this extremum is a local maximum of
`f(x,y)` with a value of 21.