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प्रश्न
For the demand function p = 550 – 3x – 6x2 where x is quantity demand and p is unit price. Show that MR =
उत्तर
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 ηd = `- "p"/x * "dx"/"dp"`
p = 550 – 3x – 6x2
`"dp"/"dx"` = 0 - 3 - 12x
∴ ηd = `- "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
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