What is the extreme f(x,y)=xy+2x-ln(x^2y) ?

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

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extreme f(x,y)=xy+2x-ln(x^2y)

extreme points

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

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

Solution steps

Find the critical points:

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

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{1}{y ^{2}}

D (x, y) = \frac{2}{x ^{2} y ^{2}} - 1

Check the sign of at each critical point

Check critical point

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

Minimum

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

2 x^{3} - 2 x^{2} - 2 x + 3

f (x) = 6 x^{2} - 12, - 4 \leq x \leq 1

f (y, x) = 0.6 x + 0.1 y - 0.01 x y

f (x) = 3 x^{4} - 8 x^{3} + 10

f (x) = x^{2} - 7 x + 12

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