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

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

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

extreme points

f (x, y) = \frac{2 x y}{( 2 x + 3 y - 4 )}

Saddle (2, \frac{4}{3}), Saddle (0, 0)

Solution steps

Find the critical points:

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

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{12 x ( 2 x - 4 )}{( 2 x + 3 y - 4 ) ^{3}}

D (x, y) = \frac{- 256 x ^{2} - 768 x y + 1024 x - 576 y ^{2} + 1536 y - 1024}{( - 4 + 2 x + 3 y ) ^{6}}

Check the sign of at each critical point

Check critical point

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

Saddle

Check critical point

(0, 0) :

Saddle

Saddle (2, \frac{4}{3}), Saddle (0, 0)

f (x) = - 2 x + 3 ln (4 x), (1, 5)

f (x) = 3 x^{2} - 30 x + 72

f (x) = x^{3} + 3 x^{2} - 9

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

f (x) = x ln (x^{2})

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