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

The extreme f(x,y)=4xy-x^3-2y^2 is Saddle(0,0),Maximum(4/3 , 4/3)

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

extreme points

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

Saddle (0, 0), Maximum (\frac{4}{3}, \frac{4}{3})

Solution steps

Find the critical points:

(0, 0), (\frac{4}{3}, \frac{4}{3})

\frac{\partial ^{2} f}{\partial x ^{2}} = - 4

D (x, y) = 24 x - 16

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(\frac{4}{3}, \frac{4}{3}) :

Maximum

Saddle (0, 0), Maximum (\frac{4}{3}, \frac{4}{3})

L (x, y) = 4 + x^{3} + y^{3} - 3 x y

f (x, y) = x^{3} + 3 y^{3} + 3 x^{2} + 3 y^{2} + 24

f (x) = 2 x + \frac{2}{x}

f (x) = \frac{( x ^{2} - 4 )}{x ^{2} + 4}

f (x, y) = x^{3} - 9 x y + y^{3}

What is the extreme f(x,y)=4xy-x^3-2y^2 ?
The extreme f(x,y)=4xy-x^3-2y^2 is Saddle(0,0),Maximum(4/3 , 4/3)