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Question
The marginal revenue function for a firm given by MR = `2/(x + 3) - (2x)/(x + 3)^2 + 5`. Show that the demand function is P = `(2x)/(x + 3)^2 + 5`
Solution
MR = `2/(x + 3) - (2x)/(x + 3)^2 + 5`
= `(2(x + 3) - 2x)/(x + 3)^2 + 5`
= `(2x + 6 - 2x)/(x + 3)^ + 5`
MR = `6/(x +3)^2 + 5`
Revenue function R = `int "MR" "dx`
= `int [6/(x + 3)^2 + 5] "d"x`
= `6int [(x + 3)^-2 + 5] "d"x`
R = `6[(x + 3)^(-2 + 1)/((-2 + 1))] + 5x + "k"`
R = `(6(x + 3)^-1)/(-1) + 5x + "k"`
R = `(-6)/((x + 3)) + 5x + "k"` .......(1)
When x = 0
R = 0
⇒ 0 = `(-6)/((0 + 3)) + 5(0) + "k"`
0 = `-2 + "k"`
⇒ k = 2
From (1)
⇒ R = `(-6)/((x + 3)) + 5x + 2`
= `2 - 6/((x + 3)) + 5x`
= `(2(x + 3) - 6)/((x + 3)) + 5x`
= `(2x + 6 - 6)/((x + 3)) + 5x`
R = `(2x)/((x + 3)) + 5x`
The demand function P = `"R"/x`
= `([(2x)/((x + 3)) + 5x])/x`
= `(x[2/(x + 3) + 5])/x`
∴ P = `2/((x + 3)) + 5`
Hence proved
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