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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

संबंधित प्रश्‍न

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A manufacturing company produces x items at a total cost of ₹ 40 + 2x. Their price is given as p = 120 – x. Find the value of x for which revenue is increasing.


A manufacturing company produces x items at a total cost of ₹ 40 + 2x. Their price is given as p = 120 – x. Find the value of x for which profit is increasing.


Find MPC, MPS, APC and APS, if the expenditure Ec of a person with income I is given as Ec = (0.0003) I2 + (0.075) I ; When I = 1000.


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The manufacturing company produces x items at the total cost of ₹ 180 + 4x. The demand function for this product is P = (240 − 𝑥). Find x for which profit is increasing


A manufacturing company produces x items at a total cost of ₹ 40 + 2x. Their price per item is given as p = 120 – x. Find the value of x for which revenue is increasing

Solution: Total cost C = 40 + 2x and Price p = 120 – x

Revenue R = `square`

Differentiating w.r.t. x,

∴ `("dR")/("d"x) = square`

Since Revenue is increasing,

∴ `("dR")/("d"x)` > 0

∴ Revenue is increasing for `square`


A manufacturing company produces x items at a total cost of ₹ 40 + 2x. Their price per item is given as p = 120 – x. Find the value of x for which profit is increasing

Solution: Total cost C = 40 + 2x and Price p = 120 − x

Profit π = R – C

∴ π = `square`

Differentiating w.r.t. x,

`("d"pi)/("d"x)` = `square`

Since Profit is increasing,

`("d"pi)/("d"x)` > 0

∴ Profit is increasing for `square`


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Ec = (0.0003)I2 + (0.075)I2

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Hence, profit is increasing for `Q < square` 


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