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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 revenue is increasing.
उत्तर
Let C be the total cost function.
∴ C = 40 + 2x
Revenue = Price × Demand
∴ `"R" = "p" × "x" = (120 - "x") * "x"`
∴ R = 120x - x2
∴ `"dR"/"dx" = 120 - 2"x" = 2(60 - "x")`
Since revenue R is an increasing function, `"dR"/"dx" > 0`
∴ 2(60 - x) > 0
∴ 60 - x > 0
∴ 60 > x
∴ x < 60
∴ The revenue R is increasing for x < 60.
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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
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∴ x = 120 – p
Differentiating w.r.t. p,
`("d"x)/("dp")` = `square`
∴ Elasticity of demand is given by η = `- "P"/x*("d"x)/("dp")`
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