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

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

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

extreme

Maximum (\frac{2}{3}, \frac{1}{3})

Solution steps

Find the critical points:

(\frac{2}{3}, \frac{1}{3})

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

D (x, y) = 3

Check the sign of at each critical point

Check critical point

(\frac{2}{3}, \frac{1}{3}) :

Maximum

Maximum (\frac{2}{3}, \frac{1}{3})

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