What is the extreme f(x)=xy+3/x+9/y ?

The extreme f(x)=xy+3/x+9/y is Minimum(1,3)

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extreme f(x)=xy+3/x+9/y

extreme points

f (x) = x y + \frac{3}{x} + \frac{9}{y}

Minimum (1, 3)

Solution steps

Find the critical points:

(1, 3)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{18}{y ^{3}}

D (x, y) = \frac{108}{x ^{3} y ^{3}} - 1

Check the sign of at each critical point

Check critical point

(1, 3) :

Minimum

Minimum (1, 3)

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

\frac{e ^{- x}}{( 1 + e ^{- x} ) ^{2}}

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

4 x^{3} - 12 x^{2}

3 x y - x^{2} y - x y^{2}

What is the extreme f(x)=xy+3/x+9/y ?
The extreme f(x)=xy+3/x+9/y is Minimum(1,3)