Question

求y=6x+2/3x^2(x>0)的最小值 已知x ∈(0,π/2),求y=(sinx)^2*cosx的最小值

Answer 1

①用到的原理为x+y+z≥3(xyz)^(1/3) (x、y、z＞0) y＝3x+3x+2/(3x^2)≥3[3x*3x*2/(3x^2)]^(1/3)＝3*6^(1/3) ∴y(min)＝3*6^(1/3) ②用到的原理为x+y+z≤[(x+y+z)/3]^3 (x、y、z＞0) y^2＝(sinx)^2*(sinx)^2*(cosx)^2＝(1/2)*[(sinx)^2*(sinx)^2*2(cosx)^2 ]≤(1/2)*{[ (sinx)^2+(sinx)^2+2(cosx)^2 ]/3}^3＝4/27 ∴y(max)＝2√3/9