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Question
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.
Solution
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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If the elasticity of demand η = 1, then demand is ______.
If 0 < η < 1, then the demand is ______.
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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 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 f(x) = x3 – 3x2 + 3x – 100, x ∈ R then f"(x) is ______.
