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)

f (x) = x^{4} + 4 x^{2} + 1

f (x) = 2 x^{3} + 9 x^{2} - 108 x + 7

150 + 8 x^{3} + x^{4}

1

f (x) = - 3 x^{4} + 4 x^{3} + 72 x^{2}

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)