# 几道英文的数学应用题. 大学经济类的.

1.An efficiency study of the morning shift at a certain factory indicates that an

average worker who arrives on the job at 8:00 am will have assembled

f (x) = −x3次方+6 x的2次方 +15x transistor radios x hours later.

a.How many radios will the worker actually assemble between 9:00 am and

10:00 am? 答案是26 不知道怎么的出来的

2.It is projected that x months from now, the population of a certain town will be P(x)=2x+4乘以x的2分之3次方+5000 million.

a. At what rate will the population be changing with respect to time 9 months

from now?

b. What will be the percentage change in population during the next six months from now? (Use differential approximation to estimate.) 答案是1.42% 老算不对.. 不应该是用 [P'(9)*6]/P(9) *100% 来算么?

3.A TV cable company has 1,000 subscribers who are each paying \$50 per month. It can get 100 more subscribers for each dollar decrease in the monthly fee.What monthly rate will yield maximum revenue and what will this revenue be? 答案是mouth rate = \$30/month, revenue = \$90,000.不大明白题目的意思..

### 1 個解答

• 最愛解答

(1) (a) From 8:00am to 9:00am, number of radios produced = -1 + 6 + 15 = 20From 8:00 am to 10:00am, number of radios produced = -8 + 24 + 30 = 46Therefore from 9:00am to 10:00am number of radios produced = 46 – 20 = 26(2)(a) P(x) = 2x + 4x^(3/2) + 5000P’(x) = 2 + 6x^(1/2)P’(6) = 2 + 6√9 = 20 million per month(b) P(6) = 2(6) + 4[6^(3/2)] + 5000 = 5070.788Percentage change = (5070.788 – 5000) / 5000 × 100% = 1.42%(3) Let the monthly rate be \$xx is (50 – x) below 50Therefore the number of subscriber is 100(50 – x) above 1000Number f subscribers = 1000 + 100(50 – x) = 6000 – 100xTotal revenue = monthly rate × number of subscribers= x(6000 – 100x)= -100(x^2 – 60x)= -100(x^2 – 60x + 900) + 90000= 90000 – 100(x – 30)^2When x = 30, the above is a maximumTherefore maximum revenue is \$90,000 when the monthly rate is \$30Alternative Revenue R(x) = 6000x – 100x^2R’(x) = 6000 – 200xR’(x) = 0 => x = 30 and R(30) = 90000

2011-03-13 21:32:42 補充：

(2)(a) should be P'(9) instead of P'(6)