Question

Given the marginal revenue function `4/(2x + 3)^2 - 1` show that the average revenue function is P = `4/(6x + 9) - 1`

Answer 1

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`