题目应该是a^2+b^2/2=1吧,
此时a*√(1+b^2)=√[a^2+(ab)^2]=√[a^2+a^2*2*(1-a^2)]=√[-2a^4+3a^2]=√[-2(a^2-3/4)^2+9/8]
故取最大值时,-2(a^2-3/4)^2=0,此时a^2=3/4,最大值=√(9/8)=3√2/4
题目应该是a^2+b^2/2=1吧,
此时a*√(1+b^2)=√[a^2+(ab)^2]=√[a^2+a^2*2*(1-a^2)]=√[-2a^4+3a^2]=√[-2(a^2-3/4)^2+9/8]
故取最大值时,-2(a^2-3/4)^2=0,此时a^2=3/4,最大值=√(9/8)=3√2/4