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प्रश्न
Find the price for the demand function D = `((2"p" + 3)/(3"p" - 1))`, when elasticity of demand is `11/14`.
उत्तर
Given, elasticity of demand (η) = `11/14` and demand function is D = `((2"p" + 3)/(3"p" - 1))`
∴ `"dD"/"dp" = (("3p" - 1)"d"/"dp" ("2p" + 3) - ("2p" + 3) "d"/"dp" ("3p" - 1))/("3p" - 1)^2`
`= (("3p" - 1)(2 + 0) - ("2p" + 3)(3 - 0))/("3p" - 1)^2`
∴ `"dD"/"dp" = (6"p" - 2 - "6p" - 9)/("3p" - 1)^2 = (- 11)/("3p" - 1)^2`
`eta = (-"p")/"D" * "dD"/"dp"`
∴ `11/14 = (-"p")/((2"p" + 3)/(3"p" - 1)) * (- 11)/("3p" - 1)^2`
∴ `11/14 = (11 "p")/(("2p" + 3)("3p" - 1))`
∴ 11 (2p + 3) (3p - 1) = 11p × 14
∴ 6p2 - 2p + 9p - 3 = 14p
∴ 6p2 + 7p - 14p - 3 = 0
∴ 6p2 - 7p - 3 = 0
∴ (2p - 3)(3p + 1) = 0
∴ 2p - 3 = 0 or 3p + 1 = 0
∴ p = `3/2` or p = `-1/3`
But, p ≠ `-1/3`
∴ p = `3/2`
∴ The price for elasticity of demand (η) = `11/14` is `3/2`.
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Solution: Total cost C = 40 + 2x and Price p = 120 − x
Profit π = R – C
∴ π = `square`
Differentiating w.r.t. x,
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`("d"pi)/("d"x)` > 0
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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`
If profit is increasing , then `(dpi)/(dQ) >0`
∴ `Q < square`
Hence, profit is increasing for `Q < square`
