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प्रश्न
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.
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उत्तर
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"]`
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