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

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

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

extreme points

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

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)

2 x + \frac{1}{x}

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

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

f (x) = x^{4} - 5 x^{3} - 15 x^{2}

f (x, y) = x + 2 y

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