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

The extreme f(x,y)=2xy-(x^2y)/3-(2xy^2)/3 is Saddle(0,3),Maximum(2,1)

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

extreme points

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

Saddle (0, 3), Maximum (2, 1)

Solution steps

Find the critical points:

(0, 3), (2, 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{4 x}{3}

D (x, y) = \frac{- 4 x ^{2} - 8 x y - 16 y ^{2}}{9} - 4 + \frac{8 ( x + 2 y )}{3}

Check the sign of at each critical point

Check critical point

(0, 3) :

Saddle

Check critical point

(2, 1) :

Maximum

Saddle (0, 3), Maximum (2, 1)

f (x) = 4 sec (x)

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

f (x, y) = x^{3} - 9 x y - 4 y^{2}

y = 4 - 6 x^{2}

\frac{x ^{2} + 2}{x - 3}

What is the extreme f(x,y)=2xy-(x^2y)/3-(2xy^2)/3 ?
The extreme f(x,y)=2xy-(x^2y)/3-(2xy^2)/3 is Saddle(0,3),Maximum(2,1)