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प्रश्न
For the demand function D = 100 – `"p"^2/2`. Find the elasticity of demand at p = 6 and comment on the results.
उत्तर
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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Differentiating w.r.t. x,
`("d"pi)/("d"x)` = `square`
Since Profit is increasing,
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Differentiating w.r.t. Q, we get
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Hence, profit is increasing for `Q < square`
