A soft drink vender at a popular beach analysed his sales records and found that if he sells x cans of soda pop in one day, his profit in dollars is given by
P(x) = -0.001x^2 + 3x - 1800
1)How many cans must he sell for maximum profit and what is his maximum profit per day?
2)If he sells 1200 cans, what is his profit?
3)To gain the same profit as in part (b) how many cans must be sell?
4)Draw a graph.
Give a full explanation please. Thanks a lot.
A Maths Question - 10 Points
2008-04-20 7:37 am
回答 (1)
2008-04-20 8:20 am
✔ 最佳答案
1) d/dx P(x) = -0.002x ＋ 3put d/dx P(x) = 0
-0.002x = -3
x = 1500
d^2/dx^2 P(x) = -0.002 which is less than 0
so P (x) attains its max. when x equals to 1500
so P (1500) = $ 450
2) P(1200) = $ 360
3) 360 = -0.001x^2 ＋ 3x - 1800
x^2 - 3000x ＋ 2160000 = 0
(x-1200)(x-1800) = 0
so x = 1200 or 1800
參考: 畫唔到圖 sor
收錄日期: 2021-04-13 15:27:28
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