设a,b的取值范围为大于0，2a*a+b*b=1，则S=2ab-4a*a*a*a-b*b*b*b的最大值？

S=2ab-4a^4-b^4
=2ab-2a^2b^2-(4a^4+b^2+4a^2b^2)
=2ab-2a^2b^2-(2a^2+b^2)^2
=2ab-2a^2b^2-1
=-(2a^2b^2-2ab+1/2)+1/2
=-(√2ab-√2/2)^2+1/2
《1/2
(当且仅当√2ab=√2/2时取等，即ab=1/2)

2a^2+b^2=1》2√2ab => ab《√2/4<1/2

所以S=2ab-4a^4-b^4
=-(√2ab-√2/2)^2+1/2
《-(√2*√2/4-√2/2）^2+1/2=(√2-1)/4

S=2ab-4a^4-b^4
=2ab+4a^2b^2-(4a^4+b^4+4a^2b^2)
=2ab+4a^2b^2-(2a^2+b^2)^2
=2ab+4a^2b^2-1
=(4a^2b^2+2ab+1/4)-5/4
=(2ab+1/2)^2-5/4

要s最大 只要ab最大 2a^2=b^2=1/2 及a=1/2 b=√2/2
S=(√2-1)/2