Question

For the demand function p x = 100 - 6x², find the marginal revenue and also show that MR = p`[1 - 1/eta_"d"]`

Answer 1

Given p = 100 - 6x²

we know that R = px

R = `(100 - 6x^2)x = 100x - 6x^3`

Marginal Revenue (MR) = `"dR"/"dx" = "d"/"dx" (100x - 6x^3)`

LHS = MR = `100 - 18x^2`   ....(1)

Differentiating p, with respect to, 'x' we get,

`"dp"/"dx" = - 12 x`

`"dx"/"dp" = (-1)/(12x)`

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

`= (- (100 - 6x^2))/x * ((-1)/(12x))`

`= (100 - 6x^2)/(12x^2)`

`therefore "RHS" = "p"(1-1/(eta_"d"))`

`= (100 - 6x^2)(1 - (12x^2)/(100 - 6x^2))`

RHS = `(cancel(100 - 6x^2))((100 - 6x^2 - 12x^2)/cancel(100 - 6x^2))`

= 100 - 18x²   ...(2)

From (1) and (2), LHS = RHS

`therefore "MR" = "p"(1 - 1/eta_"d")`