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 points

f (x) = x^{2} - 6 x y + 2 y^{2} + 10 x + 2 y - 5

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})

25 x

f (x) = x - \frac{250}{x}

f (x) = - 25 x^{2} - 25 y^{2} + 200 x + 200 y + 200

f (x) = \frac{( x + 8 )}{x ^{2}}

- x^{3} + 3 x + 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)