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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 profit is increasing. - Mathematics and Statistics

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प्रश्न

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.

बेरीज

उत्तर

Let C be the total cost function.

∴ C = 40 + 2x

Profit = Revenue - Cost

∴ π = R - C

∴ π = 120x - x2 - (40 + 2x)

= 120x - x2 - 40 - 2x

∴ π = `- "x"^2 + 118"x" - 40`

∴ `("d"pi)/"dx" = - 2"x" + 118 = 2(- "x" + 59)`

Since profit π is an increasing function, `("d"pi)/"dx" > 0`

∴ 2(- x + 59) > 0

∴ - x + 59 > 0

∴ 59 > x

∴ x < 59

∴ The profit π is increasing for x < 59.

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Application of Derivatives to Economics
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 4: Applications of Derivatives - Exercise 4.4 [पृष्ठ ११३]

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बालभारती Mathematics and Statistics 1 (Commerce) [English] 12 Standard HSC Maharashtra State Board
पाठ 4 Applications of Derivatives
Exercise 4.4 | Q 12.2 | पृष्ठ ११३

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State whether the following statement is True or False:  

If the marginal revenue is 50 and the price is ₹ 75, then elasticity of demand is 4


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 elasticity of demand for price ₹ 80.

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

p = 120 – x

∴ x = 120 – p

Differentiating w.r.t. p,

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

∴ Elasticity of demand is given by η = `- "P"/x*("d"x)/("dp")`

∴ η = `square`

When p = 80, then elasticity of demand η = `square`


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Ec = (0.0003)I2 + (0.075)I2

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In a factory, for production of Q articles, standing charges are ₹500, labour charges are ₹700 and processing charges are 50Q. The price of an article is 1700 - 3Q. Complete the following activity to find the values of Q for which the profit is increasing.

Solution: Let C be the cost of production of Q articles.

Then C = standing charges + labour charges + processing charges

∴ C = `square` 

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Profit `pi = R - C = square`

 Differentiating w.r.t. Q, we get

`(dpi)/(dQ) = square`

If profit is increasing , then `(dpi)/(dQ) >0`

∴ `Q < square` 

Hence, profit is increasing for `Q < square` 


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