What is the extreme P(x,y)=30x^2+2xy-4y^2+47x-12y+7 ?

The extreme P(x,y)=30x^2+2xy-4y^2+47x-12y+7 is Saddle(-8/11 ,-37/22)

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extreme P(x,y)=30x^2+2xy-4y^2+47x-12y+7

extreme points

P (x, y) = 30 x^{2} + 2 x y - 4 y^{2} + 47 x - 12 y + 7

Saddle (- \frac{8}{11}, - \frac{37}{22})

Solution steps

Find the critical points:

(- \frac{8}{11}, - \frac{37}{22})

\frac{\partial ^{2} f}{\partial x ^{2}} = - 8

D (x, y) = - 484

Check the sign of at each critical point

Check critical point

(- \frac{8}{11}, - \frac{37}{22}) :

Saddle

Saddle (- \frac{8}{11}, - \frac{37}{22})

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

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

f (x) = x^{\frac{3}{5}} (4 - x)

f (x) = x y + \frac{50}{x} + \frac{20}{y}

f (x, y) = (5 - x) (5 - y) (x + y - 5)

What is the extreme P(x,y)=30x^2+2xy-4y^2+47x-12y+7 ?
The extreme P(x,y)=30x^2+2xy-4y^2+47x-12y+7 is Saddle(-8/11 ,-37/22)