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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

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#### Solution

Now, Profit = Revenue − Total cost

∴ π = R − C

= 240x − x^{2} − (180 + 4x)

= 240x − x^{2} − 180 − 4x

∴ π = − x^{2} + 236x − 180

∴ `("d"pi)/("d"x)` = −2x + 236 = 2(− x + 118)

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

∴ 2(−x + 118) > 0

∴ − x + 118 > 0

∴ 118 > x

∴ x < 118

∴ The profit is increasing for x < 118.

#### RELATED QUESTIONS

Find the marginal revenue if the average revenue is 45 and elasticity of demand is 5.

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**Fill in the blank:**

A road of 108 m length is bent to form a rectangle. If the area of the rectangle is maximum, then its dimensions are _______.

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If the elasticity of demand η = 1, then demand is ______.

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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

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`

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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E_{c} = (0.0003)I^{2} + (0.075)I^{2}

when I = 1000

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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`