What is the extreme f(x,y)=5(x-3)^2-6y^2+4xy ?

The extreme f(x,y)=5(x-3)^2-6y^2+4xy is Saddle(45/17 , 15/17)

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extreme f(x,y)=5(x-3)^2-6y^2+4xy

extreme points

f (x, y) = 5 (x - 3)^{2} - 6 y^{2} + 4 x y

Saddle (\frac{45}{17}, \frac{15}{17})

Solution steps

Find the critical points:

(\frac{45}{17}, \frac{15}{17})

\frac{\partial ^{2} f}{\partial x ^{2}} = - 12

D (x, y) = - 136

Check the sign of at each critical point

Check critical point

(\frac{45}{17}, \frac{15}{17}) :

Saddle

Saddle (\frac{45}{17}, \frac{15}{17})

f (x) = x^{3} - 6 x^{2} + 4 x - 2

f (x, y) = ln (x) + ln (y) - ln (x + y) - \frac{x}{12} - \frac{y}{3}

f (x, y) = 8 x^{4} + y^{2} - 16 x y

f (x) = - 2 x - 1

f (x, y) = \frac{x y}{x + y}

What is the extreme f(x,y)=5(x-3)^2-6y^2+4xy ?
The extreme f(x,y)=5(x-3)^2-6y^2+4xy is Saddle(45/17 , 15/17)