求2t(t-2)²的最大值 【0＜t≤2】

求2t(t-2)²的最大值 【0＜t≤2】
求2t(t-2)²的最大值 【0＜t≤2】

求2t(t-2)²的最大值 【0＜t≤2】
设y = 2t(t-2)^2
最大值或最小值时,dy/dt = 0
微分得dy/dt = 4t(t-2)+2(t-2)^2
∴4t(t-2)+2(t-2)^2 = 0
4t^2 - 8t + 2(t^2 - 4t + 4) = 0
6t^2 - 16t + 8 = 0
3t^2 - 8t + 4 = 0
(3t - 2)(t - 2) = 0
t = 2/3 或 2
代进y = 2t(t-2)^2：
t = 2/3: y = 4/3(-4/3)^2 = 64/27
t = 2: y = 4(0)^2 = 0
故y的最大值是64/27,此处t = 2/3