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

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

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

extreme points

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

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

Solution steps

Find the critical points:

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

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

D (x, y) = \frac{4}{x ^{3} y ^{3}} - 64

Check the sign of at each critical point

Check critical point

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

Minimum

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

f (x, y) = 3 x y - 6 x - 3 y + 7

5 + 6 x - 8 x^{3}

y = (t - 3 t^{2})^{\frac{1}{3}}

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

f (1) = x^{3} - 6 x^{2} + 9 x + 2

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