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Goryan [66]
3 years ago
9

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

Mathematics
1 answer:
aev [14]3 years ago
5 0
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.
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