What is the extreme f(x,y)=xy+1/x+8/y ?

The extreme f(x,y)=xy+1/x+8/y is Minimum(1/2 ,4)

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extreme f(x,y)=xy+1/x+8/y

extreme points

f (x, y) = x y + \frac{1}{x} + \frac{8}{y}

Minimum (\frac{1}{2}, 4)

Solution steps

Find the critical points:

(\frac{1}{2}, 4)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{16}{y ^{3}}

D (x, y) = \frac{32}{x ^{3} y ^{3}} - 1

Check the sign of at each critical point

Check critical point

(\frac{1}{2}, 4) :

Minimum

Minimum (\frac{1}{2}, 4)

- 3 x^{\frac{5}{3}} - 15 x^{\frac{2}{3}}

x^{3} - 12 x + 5

f (x) = 4 x^{2} - 8 x + 16

y = x 9 - x^{2}

f (x) = \frac{x ^{2} - 4}{9 - x ^{2}}

What is the extreme f(x,y)=xy+1/x+8/y ?
The extreme f(x,y)=xy+1/x+8/y is Minimum(1/2 ,4)