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Tamil Nadu Board of Secondary EducationHSC Commerce Class 12

The demand equation for a product is Pd = 20 – 5x and the supply equation is Ps = 4x + 8. Determine the consumers surplus and producer’s surplus under market equilibrium

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Question

The demand equation for a product is Pd = 20 – 5x and the supply equation is Ps = 4x + 8. Determine the consumers surplus and producer’s surplus under market equilibrium

Sum
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Solution

Pd = 20 – 5x and Ps = 4x + 8

At market equilibrium

Pd = Pd

20 – 5x = 4x + 8

⇒ 20 – 8 = 4x + 5x

9x = 12

⇒ x = `12/9`

∴ x = `4/3`

When x0 = `4/3`

P0 = `20 - 5(4/3)`

= `20 - 20/3`

P0 = `(60 - 20)/3`

= `40/3`

C.S = `int_0^(x_0) "f"(x)  "d"x - x_0"p"_0`

= `int_0^(4/3) (20 - 5x)  "d"x - (4/3) (40/3)`

= `[20x - (5x^2)/2]_0^(4/3) - 160/9`

= `[20(4/3) - (5(4/3)^2)/2] - (0) - 160/9`

= `[80/3 - (5(16/9))/2] - 160/9`

= `80/3 - 80/18 - 160/9`

= `80/3 - 40/9 - 160/9`

= `(3(80) - 40 - 160)/9`

= `(240 - 200)/9`

C.S = `40/9` units

P.S = `x_0"p"_0 - int_0^(x_0) "g"(x)  "d"x`

= `(4/3) (40/3) - int_0^(4/3) (4x + 8)  "d"x`

= `160/9 - [(4x^2)/2 + 8x]_0^(4/3)`

= `160/9 - [2x^2+ 8x]_0^(4/3)`

= `60/9 -{[2(4/3)^2 +8(4/3)] - [0]}`

= `160/9 - [2(16/9) + 32/3]`

= `160/9 - 32/9 - 32/3`

= `(160 - 32 - 3(32))/9`

= `(160 - 32 - 96)/9`

= `(160- 128)/9`

∴ P.S = `32` units

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Application of Integration in Economics and Commerce
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Chapter 3: Integral Calculus – 2 - Miscellaneous problems [Page 77]

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Samacheer Kalvi Business Mathematics and Statistics [English] Class 12 TN Board
Chapter 3 Integral Calculus – 2
Miscellaneous problems | Q 6 | Page 77

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