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

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

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

extreme points

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

Minimum (3, 1)

Solution steps

Find the critical points:

(3, 1)

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

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

Check the sign of at each critical point

Check critical point

(3, 1) :

Minimum

Minimum (3, 1)

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

x^{3} + 6 x^{2} + 12 x + 2

f (x) = 27 x - x^{3}

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

f (x) = \frac{1}{x} + ln (x), e^{- 1} \leq x \leq e

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