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Tamil Nadu Board of Secondary EducationHSC Commerce Class 11

For the demand function p = 550 – 3x – 6x2 where x is quantity demand and p is unit price. Show that MR =

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Question

For the demand function p = 550 – 3x – 6x2 where x is quantity demand and p is unit price. Show that MR = 

Sum
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Solution

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- 18x

Now ηd = `- "p"/x * "dx"/"dp"`

p = 550 – 3x – 6x

`"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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Applications of Differentiation in Business and Economics
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Chapter 6: Applications of Differentiation - Exercise 6.1 [Page 139]

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Samacheer Kalvi Business Mathematics and Statistics [English] Class 11 TN Board
Chapter 6 Applications of Differentiation
Exercise 6.1 | Q 10 | Page 139

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