y=4(x^3)-40(x^2)+100x
0<x<5,如何計算Y的最大值和對應的X值?
請說明得簡易一點,謝謝
二元一次函数問題
2010-03-22 1:44 am
回答 (3)
2010-03-22 7:05 am
✔ 最佳答案
y = 4x^3 - 40x^2 + 100xdy/dx = 12x^2 - 80x + 100
Put dy/dx = 0
3x^2 - 20x + 25 = 0
(3x - 5)(x - 5) = 0
x = 5/3 or 5.
y" = 24x - 80
when x = 5/3, y" < 0 so is a max.
when x = 5, y" > 0, so is a min.
That means y is a max. when x = 5/3
and y max. = 4(5/3)(5/3 - 5)^2 = (20/3)(100/9) = 2000/27.
2010-03-23 7:58 pm
The maxcimum value=4(125)-40(25)+100(5)=0 when x=5
The minimum value=0 when x=0
Therefore, the function can not be solved.
The minimum value=0 when x=0
Therefore, the function can not be solved.
2010-03-22 4:35 am
y=4(x^3)-40(x^2)+100x
y=x(4x^2-40x+100)
y=x(2x+10)(2x+10)
y=4x(x+5)(x+5)
設x=5
y=4(5)((5)+5)((5)+5)
y=2000
x<5
y<2000
對應的X值
自己找
y=x(4x^2-40x+100)
y=x(2x+10)(2x+10)
y=4x(x+5)(x+5)
設x=5
y=4(5)((5)+5)((5)+5)
y=2000
x<5
y<2000
對應的X值
自己找
收錄日期: 2021-04-20 12:54:03
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100321000051KK01273