函数y=x√(1-2x²)的最大值(x＞0)

函数y=x√(1-2x²)的最大值(x＞0)

解由题知y＞0
故由y=x√(1-2x²)
平方得y^2=x^2(1-2x^2)
=1/2×2x^2(1-2x^2)
≤1/2×[(2x^2+(1-2x^2))/2]^2
=1/2×[1/2]^2
=1/8
故y^2≤1/8
故y≤√1/8=√1×2/8×2=√2/4
故函数y=x√(1-2x²)的最大值√2/4