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