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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 + 2

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Question

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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Sum
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Solution

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

Profit π = R – C

∴ π = 120x – x2 – (40 + 2x)

= 120x – x2 – 40 – 2x

= – x2 + 118x – 40  

Differentiating w.r.t. x,

`("d"pi)/("d"x)` = – 2x + 118 

Since Profit is increasing,

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

∴ – 2x + 118 > 0

∴ 2x < 118

∴x < 59

∴ Profit is increasing for x < 59.

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Chapter 1.4: Applications of Derivatives - Q.6

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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`


Complete the following activity to 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)I2

when I = 1000


If f(x) = x3 – 3x2 + 3x – 100, x ∈ R then f"(x) is ______.


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` 

Revenue R = P·Q = (1700 - 3Q)Q = 1700Q- 3Q2

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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