If a2x4+b2y4=c6 , then the maximum value of xy is

a2x4+b2y4=c6 ⇒y=(c6−a2x4b2)1/4 ⇒f(x)=xy=x(c6−a2x4b2)1/4 ⇒f(x)=(c6x4−a2x8b2)1/4 Differentiating f(x)  w.r.to x, then f'(x)=14(c6x4−a2x8b2)−3/4(4x3c6b2−8x7a2b2) Put f'(x)=0,4x3c6b2−8x7a2b2=0 ⇒x4=c62a2⇒x=±c3/221/4a At x=c3/221/4a,f(x)  will be maximum, so(fc3/221/4a)=(c122a2b2−c124a2b2)1/4=(c124a2b2)1/4                                      =c32ab