What is the extreme f(x,y)=-8x^2-3y^2+8x-y+2 ?

The extreme f(x,y)=-8x^2-3y^2+8x-y+2 is Maximum(1/2 ,-1/6)

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extreme f(x,y)=-8x^2-3y^2+8x-y+2

extreme

Maximum (\frac{1}{2}, - \frac{1}{6})

Solution steps

Find the critical points:

(\frac{1}{2}, - \frac{1}{6})

\frac{\partial ^{2} f}{\partial x ^{2}} = - 6

D (x, y) = 96

Check the sign of at each critical point

Check critical point

(\frac{1}{2}, - \frac{1}{6}) :

Maximum

Maximum (\frac{1}{2}, - \frac{1}{6})

What is the extreme f(x,y)=-8x^2-3y^2+8x-y+2 ?
The extreme f(x,y)=-8x^2-3y^2+8x-y+2 is Maximum(1/2 ,-1/6)