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प्रश्न
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`
उत्तर
Total cost C = 40 + 2x and Price p = 120 – x
p = 120 – x
∴ x = 120 – p
Differentiating w.r.t. p,
`("d"x)/("dp")` = – 1
∴ Elasticity of demand is given by η = `- "P"/x*("d"x)/("dp")`
∴ η = `(-"P")/(120 - "P") (-1)`
= `"p"/(120 - "P")`
When p = 80, then elasticity of demand η = `80/(120 - 80)`
= `80/40`
= 2
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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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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.
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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
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