An owner of an E-rickshaws rental company have determined that if they Charge customers Rs:x per day to rent on E-rickshaw where 100≤x≤300, then number of E-rickshaws(n),they rent per day can be shown by linear function n(x) = 4000 – 10x. If they charge Rs:100 per day or less, they will rent all their E-rickshaws. If they charge Rs:300 or more per day, they will not rent any E- rickshaws. Based on the above information,answer the following questions.(1+1+1+1+1=5) (i). Find total revenue R as a function of x. (ii). If R(x) denote the revenue, find value of x when R(x) is maximum value. (iii). At x = 360, find the revenue collected by the company. (iv). If x= 205, find the number of E-ricshaws rented per day. (v). Find the maximum revenue collected by company.
Answers
Answer:
Several unorganised setups in Delhi provide bulk charging at night by doing power theft. The e-rickshaw owners pay fixed money (Rs 100-150) for parking and charging facilities.High maintenance cost of Rs 3000 per month for owners compared to rented vehicles. An average of Rs 300 is spend by all drivers Rs. 300 per month including expenses for charging , bribes to Traffic police and parking in few areasHow long does it take to charge on average? Typically, an e-rickshaw requires 8 to 11 hours of charging.
Step-by-step explanation:
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Answer:
(i) Revenue function is R(x) = 4000x - 10x²
(ii) The maximum value is at x =200
(iii) At x = 360, the company collects a revenue of Rs. 144000
(iv) If x = 205, 1950 E-rickshaws will be rented.
(v) The maximum revenue collected by company is Rs. 400000
Step-by-step explanation:
Given the rent per day = x 100 ≤ x ≤ 300
Number of rickshaws rented per day = n
n(x) = 4000 - 10x
It is also given that if x ≤ 100 n = n
if x ≥ 300 n = 0
(i) We know that Revenue function R(x) = Quantity * Selling price
= (4000 - 10x) * x
= 4000x - 10x²
Therefore, revenue function is R(x) = 4000x - 10x²
(ii) To find the maxima of R(x),
R'(x) = 4000 - 20x
Taking R'(x) = 0
=> 4000 - 20x = 0
=> 4000 = 20x
=> x = 200
R''(x) = 0 - 20 < 0 Therefore maximum
Therefore, the maximum value is at x =200
(iii) We need to find the revenue at x = 360
Therefore, substituting the value of x in R(x)
=> R(x) = 4000*360 - 10 * (360)²
= 360(4000 - 10*360)
= 360(4000 - 3600)
= 360 * 400
= 144000
Therefore, at x = 360, the company collects a revenue of Rs. 144000
(iv) If x = 205, we need to find n(x),
Substituting the values in n(x) = 4000 - 10x
n = 4000 - 10*205
n = 4000 - 2050
n = 1950
Therefore, if x = 205, 1950 E-rickshaws will be rented
(v) As we have proved it part (ii) that there will be maxima at x = 200
Therefore, maximum revenue = R(200)
=> R(200) = 4000*200 - 10 * (200)²
= 200(4000 - 10*200)
= 200(4000 - 2000)
= 200 * 2000
= 400000
Therefore, the maximum revenue collected by company is Rs. 400000