What is the extreme f(x,y)=-3x^2-2y^2+3x-4y+5 ?

The extreme f(x,y)=-3x^2-2y^2+3x-4y+5 is Maximum(1/2 ,-1)

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extreme f(x,y)=-3x^2-2y^2+3x-4y+5

extreme

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

Solution steps

Find the critical points:

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

\frac{\partial ^{2} f}{\partial x ^{2}} = - 4

D (x, y) = 24

Check the sign of at each critical point

Check critical point

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

Maximum

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

What is the extreme f(x,y)=-3x^2-2y^2+3x-4y+5 ?
The extreme f(x,y)=-3x^2-2y^2+3x-4y+5 is Maximum(1/2 ,-1)