超簡單微積分最大化問題(急)!!!
2011-12-17 12:29 am
一間賣電筒公司, 如果每個電筒賣 $6, 一個月可以賣出3000個電筒。價格每增加$1,每個月電筒會少賣1000個(彈性)。每個電筒成本為$4。請問每個電筒要賣幾錢才會有最大利潤(profit)?
回答 (2)
2011-12-17 5:03 am
✔ 最佳答案
Let Pirce = $x.Then number = 3000+1000(6-x)= 9000-1000x
Profit = (9000-1000x)(x-4)
d(Profit)/d(Price) = -1000(x-4)+9000-1000x= 13000-2000x
Put d(Profit)/d(Price) =0,
x = 6.5
Note that d(Profit)/d(Price)>0 at x<6.5
and<0 at x>6.5
Thus Profit is a maximum at x=6.5
Hence the price is $6.5
其實不必用calculus
用中二 二次方程 已足夠
2011-12-17 5:19 am
sorry for english only
Let x be selling price
number sold: [3000 - 1000(x-6)]
profit: [x - 4]
total profit, P = [3000 - 1000(x-6)] [x - 4]
= [3000 - 1000x + 6000] [x - 4]
= [9000 - 1000x] [x - 4]
= -1000[x - 9] [x - 4]
= -1000[x^2 - 13x + 36]
= -1000[x^2 - 13x + (13/2)^2 - (13/2)^2 + 36]
= -1000{[x - (13/2)]^2 - 169/4 + 36}
= -1000{[x - (13/2)]^2 - 25/4}
= -1000[x - (13/2)]^2 + 6250
as -1000[x - (13/2)]^2 is always < 0, profit is maximised 6250 when x = 13/2
Let x be selling price
number sold: [3000 - 1000(x-6)]
profit: [x - 4]
total profit, P = [3000 - 1000(x-6)] [x - 4]
= [3000 - 1000x + 6000] [x - 4]
= [9000 - 1000x] [x - 4]
= -1000[x - 9] [x - 4]
= -1000[x^2 - 13x + 36]
= -1000[x^2 - 13x + (13/2)^2 - (13/2)^2 + 36]
= -1000{[x - (13/2)]^2 - 169/4 + 36}
= -1000{[x - (13/2)]^2 - 25/4}
= -1000[x - (13/2)]^2 + 6250
as -1000[x - (13/2)]^2 is always < 0, profit is maximised 6250 when x = 13/2
收錄日期: 2021-04-16 17:05:23
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20111216000051KK00354