What is the extreme f(x,y)=(16)/x+6/y+x^2-3y^2 ?

The extreme f(x,y)=(16)/x+6/y+x^2-3y^2 is Saddle(2,-1)

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extreme f(x,y)=(16)/x+6/y+x^2-3y^2

extreme points

f (x, y) = \frac{16}{x} + \frac{6}{y} + x^{2} - 3 y^{2}

Saddle (2, - 1)

Solution steps

Find the critical points:

(2, - 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 6 + \frac{12}{y ^{3}}

D (x, y) = (\frac{32}{x ^{3}} + 2) (\frac{12}{y ^{3}} - 6)

Check the sign of at each critical point

Check critical point

(2, - 1) :

Saddle

Saddle (2, - 1)

f (x, y) = x^{3} - 27 x y + 27 y^{3}

f (x) = \frac{x ^{2} - 14 x + 49}{x - 9}

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

f (x) = x y^{2} - 8 x y

\frac{5 - x}{2 x + 6}

What is the extreme f(x,y)=(16)/x+6/y+x^2-3y^2 ?
The extreme f(x,y)=(16)/x+6/y+x^2-3y^2 is Saddle(2,-1)