令 f(x) = 2x^3-3(a+1)x^2+6ax ,
则 f'(x) = 6x^2-6(a+1)x+6a = 6(x-1)(x-a) ,
可得：f(x)在 x = 1 和 x = a 时取得极值.
f(1) = 2-3(a+1)+6a = 3a-1 ；
f(a) = 2a^3-3(a+1)a^2+6a^2 = (3-a)a^2 .
① 假设 f(1) 为最小值,则有：3a-1 = 4 ,解得：a = 5/3 ,
此时 f(a) = f(5/3) = 100/27 < 4 = f(1) ,这和假设 f(1) 为最小值矛盾,舍去；
② 假设 f(a) 为最小值,则有：(3-a)a^2 = 4 ,解得：a = 2 ,
此时 f(1) = 5 > 4 = f(a) ,满足假设；
综上可得：a = 2 .