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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)
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Solution
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
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