Question 955351
 {{{a^2 x^4 + b^2 y^4 = c^6}}}
{{{b^2y^4=c^6-a^2x^4}}}
{{{y^4=(c^6-a^2x^4)/b^2}}}
{{{y=((c^6-a^2x^4)/b^2)^(1/4)}}}
So then,
{{{Z=xy}}}
{{{Z=x((c^6-a^2x^4)/b^2)^(1/4)}}}
Taking the derivative of Z with respect to x,
{{{dZ/dx=(c^6-2a^2x^4)/(b^2((c^6-a^2x^4)/b^2)^(3/4))}}}
To find the extrema, set the derivative equal to zero.
{{{c^6-2a^2x^4=0}}}
{{{2a^2x^4=c^6}}}
{{{x^4=c^6/(2a^2)}}}
So then,
{{{a^2(c^6/(2a^2))+b^2y^4=c^6}}}
{{{c^6/2+b^2y^4=c^6}}}
{{{b^2y^4=c^6/2}}}}
{{{y^4=c^6/(2b^2)}}}
So then,
{{{x^4y^4=(c^6/(2a^2))(c^6/(2b^2))}}}
{{{(xy)^4=c^12/(4a^2b^2)}}}
and finally,
{{{xy[max]=(c^12^(1/4))/(4a^2b^2)^(1/4)}}}
{{{highlight(xy[max]=c^3/sqrt(2ab))}}}