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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`

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#### Solution

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

Revenue R = **px**

= (120 – x)x

∴ R = 120 – x^{2}

Differentiating w.r.t. x,

`("dR")/("d"x)` = **120 – x**

Since Revenue is increasing,

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

∴ 120 – 2x > 0

∴ 120 > 2x

∴ x < 60

∴ Revenue is increasing for **x < 60**

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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 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`

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`

If 0 < η < 1 then the demand 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- 3Q^{2 }

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`