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महाराष्ट्र राज्य शिक्षण मंडळएचएससी वाणिज्य (इंग्रजी माध्यम) इयत्ता १२ वी

Find the price for the demand function D = (2p+33p-1), when elasticity of demand is 1114. - Mathematics and Statistics

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प्रश्न

Find the price for the demand function D = `((2"p" + 3)/(3"p" - 1))`, when elasticity of demand is `11/14`.

बेरीज
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उत्तर

Given, elasticity of demand (η) = `11/14` and demand function is D = `((2"p" + 3)/(3"p" - 1))`

∴ `"dD"/"dp" = (("3p" - 1)"d"/"dp" ("2p" + 3) - ("2p" + 3) "d"/"dp" ("3p" - 1))/("3p" - 1)^2`

`= (("3p" - 1)(2 + 0) - ("2p" + 3)(3 - 0))/("3p" - 1)^2`

∴ `"dD"/"dp" = (6"p" - 2 - "6p" - 9)/("3p" - 1)^2 = (- 11)/("3p" - 1)^2`

`eta = (-"p")/"D" * "dD"/"dp"`

∴ `11/14 = (-"p")/((2"p" + 3)/(3"p" - 1)) * (- 11)/("3p" - 1)^2`

∴ `11/14 = (11 "p")/(("2p" + 3)("3p" - 1))`

∴ 11 (2p + 3) (3p - 1) = 11p × 14

∴ 6p2 - 2p + 9p - 3 = 14p

∴ 6p2 + 7p - 14p - 3 = 0

∴ 6p2 - 7p - 3 = 0 

∴ (2p - 3)(3p + 1) = 0

∴ 2p - 3 = 0    or  3p + 1 = 0

∴ p = `3/2`  or p = `-1/3`

But, p ≠ `-1/3`

∴ p = `3/2`

∴ The price for elasticity of demand (η) = `11/14` is `3/2`.

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पाठ 4: Applications of Derivatives - Exercise 4.4 [पृष्ठ ११३]

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बालभारती Mathematics and Statistics 1 (Commerce) [English] Standard 12 Maharashtra State Board
पाठ 4 Applications of Derivatives
Exercise 4.4 | Q 9 | पृष्ठ ११३

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