Question

Show that MR = p`[1 - 1/eta_"d"]` for the demand function p = 400 – 2x – 3x² where p is unit price and x is quantity demand.

Answer 1

Given p = 400 – 2x – 3x²

Revenue, R = px = (400 – 2x – 3x²)x = 400x – 2x² – 3x³

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

`= "d"/"dx"` (400x – 2x² – 3x³)

= 400 – 4x – 9x²

Now `eta_"d" = - "p"/x * "dx"/"dp"`

p = 400 - 2x - 3x²

`"dp"/"dx"` = 0 - 2 - 6x

`therefore eta_"d" = - "p"/x * 1/(("dp"/"dx"))`

`= - [(400 - 2x - 3x^2)/x] xx 1/(- 2 - 6x)`

`= (400 - 2x - 3x^2)/(-x) xx 1/(- 2 -6x)`

`= (400 - 2x - 3x^2)/(2x + 6x^2)`

`therefore 1 - 1/eta_"d" = 1 - 1/(((400 - 2x - 3x^2)/(2x + 6x^2)))`

`= 1 - ((2x + 6x^2)/(400 - 2x - 3x^2))`

`= (400 - 2x - 3x^2 - (2x + 6x^2))/(400 - 2x - 3x^2)`

`= (400 - 4x - 9x^2)/(400 - 2x - 3x^2)`

`therefore "p"[1 - 1/eta_"d"] = (400 - 2x - 3x^2)((400 - 4x - 9x^2)/(400 - 2x - 3x^2))`

= 400 - 4x - 9x² = MR  (using (1))

Thus for the function p = 400 – 2x – 3x²

MR = `"p"[1 - 1/eta_"d"]`