最大值與最小值應用?

2017-08-01 11:22 am

回答 (2)

2017-08-01 1:02 pm
✔ 最佳答案
Sol
(1)
設增加x人收入最高
f(x)=(x+26)(4000-100x)
=4000x-100x^2+104000-2600x
=-100x^2+1400x+104000
=-100(x^2-14x)+104000
=-100(x^2-14x+49)+108900
=-100(x-7)^2+108900
(x-7)^>=0
-(x-7)^2<=0
-100(x-7)^2<=0
-100(x-7)^2+108900<=108900
f(x)<=108900
x=7時
增加7人收入最高
(2) 108900元
2017-08-01 1:04 pm
每車增加 x 人時, 可收到
(26+x)(4000-100x)
=100(26+x)(40-x)
=100(-x^2+14x+1040)
=-100(x^2-14x-1040)
=-100[(x-7)^2-1089]
Ans: 當增加7人時, 每車可收到 108900 元


收錄日期: 2021-04-30 22:24:44
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20170801032215AAU3QTW

檢視 Wayback Machine 備份