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

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

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

extreme

Maximum (\frac{1}{4}, \frac{1}{4})

Solution steps

Find the critical points:

(\frac{1}{4}, \frac{1}{4})

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

D (x, y) = 8

Check the sign of at each critical point

Check critical point

(\frac{1}{4}, \frac{1}{4}) :

Maximum

Maximum (\frac{1}{4}, \frac{1}{4})

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