Question

The marginal revenue (in thousands of Rupees) functions for a particular commodity is `5 + 3"e"^(- 003x)` where x denotes the number of units sold. Determine the total revenue from the sale of 100 units. (Given e^–3 = 0.05 approximately)

Answer 1

The marginal Revenue (in thousands of Rupees) function

M.R = `5 + 3"e"^(- 003x)`

Total Revenue from sale of 100 units is

Total Revenue T.R = `int_0^100 "M.R"  "d"x`

= `int_0^100 (5 + 3"e"^(- 0.03x))  "d"x`

= `[5x + 3 (["e"^(-0.03x)])/(- 0.03)]_0^100`

= `{5x - 3 (["e"^(-0.03x)])/((3/100))}_0^100`

= `[5 x - 100 "e"^(- 0.03x)]_0^100`

= `[5(100) - 100"e"^(- 0.03(0))] - [5(0) - 100"e"^(-0.030(0))]`

= [500 – 100 e^–3] – [0 – 100 e°]

= [500 -100 (0.05)] – [– 100 (1)]

= [500 – 5]+ 100

= 495 + 100

= 595 thousands

= 595 × 1000

∴ Revenue R = ₹ 595000