Question

The marginal revenue function for a firm given by MR = `2/(x + 3) - (2x)/(x + 3)^2 + 5`. Show that the demand function is P = `(2x)/(x + 3)^2 + 5`

Answer 1

MR = `2/(x + 3) - (2x)/(x + 3)^2 + 5`

= `(2(x + 3) - 2x)/(x + 3)^2 + 5`

= `(2x + 6 - 2x)/(x + 3)^ + 5`

MR = `6/(x +3)^2 + 5`

Revenue function R = `int "MR"  "dx`

= `int [6/(x + 3)^2 + 5]  "d"x`

= `6int [(x + 3)^-2 + 5]  "d"x`

R = `6[(x + 3)^(-2 + 1)/((-2 + 1))] + 5x + "k"`

R = `(6(x + 3)^-1)/(-1) + 5x + "k"`

R = `(-6)/((x + 3)) + 5x + "k"`  .......(1)

When x = 0

R = 0

⇒ 0 = `(-6)/((0 + 3)) + 5(0) + "k"`

0 = `-2 + "k"`

⇒ k = 2

From (1)

⇒ R = `(-6)/((x + 3)) + 5x + 2`

= `2 - 6/((x + 3)) + 5x`

= `(2(x + 3) - 6)/((x + 3)) + 5x`

= `(2x + 6 - 6)/((x + 3)) + 5x`

R = `(2x)/((x + 3)) + 5x`

The demand function P = `"R"/x`

= `([(2x)/((x + 3)) + 5x])/x`

= `(x[2/(x + 3) + 5])/x`

∴ P = `2/((x + 3)) + 5`

Hence proved