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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 also find an elasticity of demand for price 80.
उत्तर
Given, the price is p = 120 - x
∴ x = 120 - p
where, x = demand
∴ `"dx"/"dp" = 0 - 1 = - 1`
`eta = (-"p")/"x" * "dx"/"dp"`
∴ `eta = (-"p")/(120 - "p") * (- 1)`
∴ `eta = "p"/(120 - "p")`
p = 80 ....(Given)
∴ `eta = 80/(120 - 80) = 80/40 = 2`
∴ The elasticity of demand for p = 80 is η = 2.
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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`
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`
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If 0 < η < 1 then the demand is ______.
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Then C = standing charges + labour charges + processing charges
∴ C = `square`
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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`
