l U Mobile 3G 6:17 PM 2796 Back Assignment 2.docx 1. A TV Shop sells 400 sets pe
ID: 3148733 • Letter: L
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l U Mobile 3G 6:17 PM 2796 Back Assignment 2.docx 1. A TV Shop sells 400 sets per month at a price of RM600 per unit. Sale increases by one additional set for each RMI reduction in the price. Let x be the number of additional sets beyond the original 400 units. a) Fill up the table below Price Unit sold Total revenue 600-1 4001 600-2 400-2 (600×400> (600-1M400 1) (600-21 1400-2) 600- b) Find the total revenue equation in terms of x c) Find the number of additional sets beyond the original 400 units that maximnesthe total revenue d) Confirm that the solution in (cwill maximize the total revenue. e) Find the total units sold ) Find the price that maximires the total revenuc. g) Find the maximum dotal revenue.Explanation / Answer
a) Price: 600-x
Unit sold = 400+x
Total revenue = (600-x)(400+x)
b) Total revenue = (600-x)(400+x)
= 240000 + 600x - 400x - x2
= - x2 + 200x + 240000
Revenue equation in terms of x: -x2 + 200x + 240000
c) to maximize the revenue: we differentiate the revenue equation with respect to x and find the corresponding x (additional units sold)
d/dx(-x2 + 200x + 240000) = 0
-2x + 200 = 0
x = 100
Hence, additional beyond 400 required to be sold to maximize revenue = 100
d) revenue at price 600 and unit sold = 400
R = 600*400 = 240000
Revenue at 99 additional units sold = (600-99)(400+99) = 501*499 = 249999
revenue at optimum 100 aditional units = (600-100)(400+100) = 500*500 = 250000
revenue at 101 additional units = (600-101)(400+101) = 499*501 = 249999
Hence we can see that the revenue is higher than the base revenue of 240000, and as the units sold increases the revenue increases, upto 100 units.
At 99 units revenue = 249999, at 100 units 250000. and again at 101 units it comes down to 249999. hence 100 additional units provide maximum revenue of 250000.
e) total units sold = 400+100 = 500 units
f) Price that maximizes total revenue = 600-100 = RM500
g) maximum revenue = 500*500 = RM250000
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