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Question
For the demand function D = 100 – `"p"^2/2`. Find the elasticity of demand at p = 6 and comment on the results.
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Solution
Given, demand function is D = 100 - `"p"^2/2`
∴ `"dD"/"dp" = 0 - "2p"/2 = -"p"`
`eta = (-"p")/"D" * "dD"/"dp"`
∴ `eta = (-"p")/(100 - "p"^2/2) * (- "p")`
`= "p"^2/((200 - "p"^2)/2)`
∴ `eta = "2p"^2/(200 - "p"^2)`
When p = 6,
`eta = (2(6)^2)/(200 - (6)^2) = 72/164 = 18/41`
∴ elasticity of demand at p = 6 is `18/41`
Here, η > 0
∴ The demand is inelastic.
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Fill in the blank:
A road of 108 m length is bent to form a rectangle. If the area of the rectangle is maximum, then its dimensions are _______.
If 0 < η < 1, then the demand is ______.
If the average revenue is 45 and elasticity of demand is 5, then marginal revenue is ______.
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 ______.
If 0 < η < 1 then the demand is ______.
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`
Revenue R = P·Q = (1700 - 3Q)Q = 1700Q- 3Q2
Profit `pi = R - C = square`
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`
