What is the extreme U(x,y)=(x+x/y)(2y+3) ?

The extreme U(x,y)=(x+x/y)(2y+3) is Saddle(0,-1),Saddle(0,-3/2)

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extreme U(x,y)=(x+x/y)(2y+3)

extreme points

U (x, y) = (x + \frac{x}{y}) (2 y + 3)

Saddle (0, - 1), Saddle (0, - \frac{3}{2})

Solution steps

Find the critical points:

(0, - 1), (0, - \frac{3}{2})

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{6 x}{y ^{3}}

D (x, y) = - \frac{9}{y ^{4}} + \frac{12}{y ^{2}} - 4

Check the sign of at each critical point

Check critical point

(0, - 1) :

Saddle

Check critical point

(0, - \frac{3}{2}) :

Saddle

Saddle (0, - 1), Saddle (0, - \frac{3}{2})

- e^{- x} x

\frac{2680}{3725}

f (x) = 3 x^{2} - 6 x + 5

f (x) = 3 x^{2} - 6 x + 9

Y (s) = \frac{α ^{1}}{s} + \frac{α ^{2}}{2 s + 6}

What is the extreme U(x,y)=(x+x/y)(2y+3) ?
The extreme U(x,y)=(x+x/y)(2y+3) is Saddle(0,-1),Saddle(0,-3/2)