What is the extreme f(x,y)=2x^3-y^2-6xy+23 ?

The extreme f(x,y)=2x^3-y^2-6xy+23 is Maximum(-3,9),Saddle(0,0)

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extreme f(x,y)=2x^3-y^2-6xy+23

extreme points

f (x, y) = 2 x^{3} - y^{2} - 6 x y + 23

Maximum (- 3, 9), Saddle (0, 0)

Solution steps

Find the critical points:

(- 3, 9), (0, 0)

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

D (x, y) = - 24 x - 36

Check the sign of at each critical point

Check critical point

(- 3, 9) :

Maximum

Check critical point

(0, 0) :

Saddle

Maximum (- 3, 9), Saddle (0, 0)

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

f (x) = 2 x^{3} + 3 x^{2} - 12 x - 18

f (x, y) = 2 x y + \frac{1}{x} + \frac{4}{y}

f (x) = 4 x^{3} e^{- x}

f (x) = x^{3} + y^{3} - 36 x y

What is the extreme f(x,y)=2x^3-y^2-6xy+23 ?
The extreme f(x,y)=2x^3-y^2-6xy+23 is Maximum(-3,9),Saddle(0,0)