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

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

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

extreme points

f (x, y) = - 3 x^{2} - 3 y^{2} + x - 3 y + 2

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

Solution steps

Find the critical points:

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

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

D (x, y) = 36

Check the sign of at each critical point

Check critical point

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

Maximum

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

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