Find the maximum value of the function fxylnxy2 subject to 2

Solution

y^2=4-x^2 So f(x,y)=ln(x(4-x^2))=ln(x)+ln(4-x^2) df/dx=1/x-2x/(4-x^2)=(4-3x^2)/(x(4-x^2)) df/dx=0 for 4-3x^2=0 x=sqrt(4/3) y=sqrt(8/3) df/dx<0 up to sqrt(4/3) and >0 after it, so at sqrt(4/3) there is a local maximum limit when x->0 f(x,y)=limit when y->0=-infinity, so at sqrt(4/3) there is a global maximum f(sqrt(4/3),sqrt(8/3))=ln(sqrt(4/3)*8/3)