What is the extreme f(x)=x^2-6xy+2y^2+10x+2y-5 ?

The extreme f(x)=x^2-6xy+2y^2+10x+2y-5 is Saddle(13/7 , 16/7)

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extreme f(x)=x^2-6xy+2y^2+10x+2y-5

extreme

Saddle (\frac{13}{7}, \frac{16}{7})

Solution steps

Find the critical points:

(\frac{13}{7}, \frac{16}{7})

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

D (x, y) = - 28

Check the sign of at each critical point

Check critical point

(\frac{13}{7}, \frac{16}{7}) :

Saddle

Saddle (\frac{13}{7}, \frac{16}{7})

What is the extreme f(x)=x^2-6xy+2y^2+10x+2y-5 ?
The extreme f(x)=x^2-6xy+2y^2+10x+2y-5 is Saddle(13/7 , 16/7)