The demand function is D=(P 3 p 1 where d demand P price find the elasticity of demand when p 5)

If the demand function is D = 50 – 3p – p2. Find the elasticity of demand at p = 5 comment on the result

Given, demand function is D = 50 – 3p – p2 

∴ `"dD"/"dp" = 0 - 3 - 2"p"`

= `- 3 - 2"p"`

`eta = (-"p")/"D" * "dD"/"dp"`

∴ `eta = (-"p")/(50 - 3"p" - "p"^2) * (- 3 - 2"p")`

∴ `eta = (3"p" + 2"p"^2)/(50 - "3p" - "p"^2)`

When p = 5

`eta = (3(5) + 2(5)^2)/(50 - 3(5) - (5)^2)`

= `(15 + 50)/(50 - 15 - 25)`

= `65/10`

∴ η = 6.5

∴ elasticity of demand at p = 5 is 6.5

Here,  η > 0

∴ The demand is elastic.

Concept: Application of Derivatives to Economics

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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`

`= (("p" - 3)(1 + 0) - ("p" + 6)(1 - 0))/("p" - 3)^2`

∴ `"dD"/"dp" = ("p" - 3 - "p" - 6)/("p" - 3)^2`

`= (-9)/("p" - 3)^2`

`eta = (-"p")/"D" * "dD"/"dp"`

∴ `eta = (- "p")/((("p" + 6)/("p" - 3))) * (-9)/("p" - 3)^2`

∴ `eta = (9"p")/(("p" + 4)("p" - 3))`

Substituting p = 4, we get

`eta = (9 xx 4)/((4 + 6)(4 - 3)) = 36/(10(1))` 

∴ η = 3.6

∴ elasticity of demand at p = 4 is 3.6

Page 2

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`.