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 points

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

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)

f (x, y) = ∣ x y ∣

f (x) = x^{3} - \frac{3}{2} x^{2}, - 3 \leq x \leq 2

f (x) = 5 - 2 x

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

f (x, y) = 5 x^{4} - x^{2} + 6 y^{2}

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)