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

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

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

extreme

Minimum (1, 1), Minimum (- 1, - 1)

Solution steps

Find the critical points:

(1, 1), (- 1, - 1)

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

D (x, y) = \frac{4}{x ^{2} y ^{2}} + \frac{2}{x ^{2}} + \frac{2}{y ^{2}}

Check the sign of at each critical point

Check critical point

(1, 1) :

Minimum

Check critical point

(- 1, - 1) :

Minimum

Minimum (1, 1), Minimum (- 1, - 1)

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