What is the extreme f(x,y)=ln(xy-x^3-y^3+x^2y^2) ?

The extreme f(x,y)=ln(xy-x^3-y^3+x^2y^2) is Maximum(1/2 , 1/2)

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extreme f(x,y)=ln(xy-x^3-y^3+x^2y^2)

extreme points

f (x, y) = ln (x y - x^{3} - y^{3} + x^{2} y^{2})

Maximum (\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 x ^{5} - 2 x ^{4} y ^{2} + 4 x ^{3} y + 4 x ^{2} y ^{3} - x ^{2} - 3 y ^{4}}{( x y - x ^{3} - y ^{3} + x ^{2} y ^{2} ) ^{2}}

D (x, y) = \frac{6 x ^{9} - 6 x ^{8} y ^{2} - 4 x ^{7} y ^{4} - 20 x ^{7} y + 4 x ^{6} y ^{6} + 28 x ^{6} y ^{3} - x ^{6} - 4 x ^{5} y ^{5} + 30 x ^{5} y ^{2} - 4 x ^{4} y ^{7} - 51 x ^{4} y ^{4} + 28 x ^{3} y ^{6} - 16 x ^{3} y ^{3} - 6 x ^{2} y ^{8} + 30 x ^{2} y ^{5} + x ^{2} y ^{2} - 20 x y ^{7} + 6 y ^{9} - y ^{6}}{( - x ^{3} - y ^{3} + x ^{2} y ^{2} + x y ) ^{4}}

Check the sign of at each critical point

Check critical point

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

Maximum

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

What is the extreme f(x,y)=ln(xy-x^3-y^3+x^2y^2) ?
The extreme f(x,y)=ln(xy-x^3-y^3+x^2y^2) is Maximum(1/2 , 1/2)