Question

For the demand function p = 550 3x 6x2where x is quantity demand and p is unit price. Show that MR =

Answer 1

Given p = 550 3x 6x2

Revenue, R = px = (550 3x 6x2)x = 550x 3x2 6x3

Marginal Revenue (MR) = $"d"/"dx"$(R)

$= "d"/"dx"$(550x 3x2 6x3)

= 550 6x- 18x2

Now d = $- "p"/x * "dx"/"dp"$

p = 550 3x 6x2

$"dp"/"dx"$ = 0 - 3 - 12x

d = $- "p"/x * 1/("dp"/"dx")$

$= - [(550 - 3x - 6x^2)/x] xx 1/((- 3 - 12x))$

$= (550 - 3x - 6x^2)/(-x) xx 1/((- 3 - 12x))$

$= (550 - 3x - 6x^2)/(3x + 12x^2)$

$therefore 1 - 1/eta_"d" = 1 - 1/(((550 - 3x - 6x^2)/(3x + 12x^2)))$

$= 1 - (3x + 12x^2)/(550 - 3x - 6x^2)$

$= (550 - 3x - 6x^2 - 3x - 12x^2)/(550 - 3x - 6x^2)$

$= (550 - 6x - 18x^2)/(550 - 3x - 6x^2)$

$therefore "p"[1 - 1/eta\_"d"] = (((550 - 3x - 6x^2)(550 - 6x - 18x^2))/(550 - 3x - 6x^2))$

$= 550 - 6x - 18x^2$ = MR