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Question
Given the marginal revenue function `4/(2x + 3)^2 - 1` show that the average revenue function is P = `4/(6x + 9) - 1`
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Solution
M.R = `4/(2x + 3)^2 - 1`
Total Revenue R =`int"M.R" "d"x`
R = `int (4/(2x + 3)^2 - 1) "d"x`
= `int [4(2x + 3)^-2 - 1] "d"x`
R = `[4[(2x + 3)^(-2 + 1)/(-2 + 1)] - x] + "k"`
R = `4[(2x + 3)^1/((-1) xx 2)] - x + "k"`
R = `4[1/(-2(2x + 3))] - x + "k"`
R = `(-2)/((2x + 3)) - x + "k"` .......(1)
When x = 0
R = 0
⇒ 0 = `(-2)/([2(0) + 3]) - 0 + "k"`
0 = `(-2)/3 + "k"``
⇒ k = `2/3`
From eqaution (1)
⇒ R = `(-2)/((2x + 3)) - x + 2/3`
⇒ R = `2/3 - 2/((2x + 3)) - x`
R = `(2(2x + 3) - 2(3))/(3(2x + 3)) - x`
= `(4x + 6 - 6)/((6x + 9)) - x`
∴ R = `(4x)/((6x + 9)) - x`
Average Revenue A.R = `"R"/x`
= `([(4x)/((6x + 9)) x])/x`
A.R = `4/((6x + 9)) - 1`
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