What is the extreme P(x,y)=15x^2+14xy+3y^2+41x+23y+14 ?

The extreme P(x,y)=15x^2+14xy+3y^2+41x+23y+14 is Saddle(-19/4 , 29/4)

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extreme P(x,y)=15x^2+14xy+3y^2+41x+23y+14

extreme points

P (x, y) = 15 x^{2} + 14 x y + 3 y^{2} + 41 x + 23 y + 14

Saddle (- \frac{19}{4}, \frac{29}{4})

Solution steps

Find the critical points:

(- \frac{19}{4}, \frac{29}{4})

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

D (x, y) = - 16

Check the sign of at each critical point

Check critical point

(- \frac{19}{4}, \frac{29}{4}) :

Saddle

Saddle (- \frac{19}{4}, \frac{29}{4})

8 x^{2} - 4 x^{4}

f (x) = \frac{5 x}{e ^{2 x^{2} + 3}}

+ x^{2} + 4 x + 6

f (x) = x^{\frac{1}{3}} + b x^{\frac{4}{3}}

(1 + x y) (x + y)

What is the extreme P(x,y)=15x^2+14xy+3y^2+41x+23y+14 ?
The extreme P(x,y)=15x^2+14xy+3y^2+41x+23y+14 is Saddle(-19/4 , 29/4)