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Question
If the demand function is D = `((p + 6)/(p − 3))`, find the elasticity of demand at p = 4.
Solution
Given, demand function is D = `((p + 6)/(p − 3))`
∴ `"dD"/"dp" = ((p − 3) "d"/"dp" (p + 6) − (p + 6) "d"/"dp" (p − 3))/(p − 3)^2`
∴ `"dD"/"dp" = ((p − 3)(1 + 0) − (p + 6)(1 − 0))/(p − 3)^2`
∴ `"dD"/"dp" = (cancelp − 3 − cancelp − 6)/(p − 3)^2`
∴ `"dD"/"dp" = (-9)/(p − 3)^2`
`eta = (−p)/"D". "dD"/"dp"`
∴ `eta = (− p)/(((p + 6)/cancel(p − 3))) . (− 9)/(p − 3)^cancel2`
∴ `eta = (9p)/((p + 6)(p - 3))`
Substituting p = 4, we get,
∴ `eta = (9 xx 4)/((4 + 6)(4 - 3))`
∴ `eta = 36/(10 × 1)`
∴ `eta = 36/10`
∴ η = 3.6
∴ η (elasticity of demand at p = 4) = 3.6.
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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
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Profit π = R – C
∴ π = `square`
Differentiating w.r.t. x,
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Since Profit is increasing,
`("d"pi)/("d"x)` > 0
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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 elasticity of demand for price ₹ 80.
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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")`
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Then C = standing charges + labour charges + processing charges
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Profit `pi = R - C = square`
Differentiating w.r.t. Q, we get
`(dpi)/(dQ) = square`
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∴ `Q < square`
Hence, profit is increasing for `Q < square`
