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 $f_{xx}(-3,-3)=-2$, this extremum is a local maximum of $f(x,y)$ with a value of 21.