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Question
The total revenue function for a commodity is R `= 15x + x^2/3 - 1/36 x^4`. Show that at the highest point average revenue is equal to the marginal revenue.
Solution
R = `15x + x^2/3 - 1/36 x^4`
Average Revenue = AR = `"R"/x`
`= (15x + x^2/3 - 1/36x^4)/x`
`= 15 + x/3 - 1/36 x^3`
To test maxima or minima for AR = `("d"("AR"))/"dx"`
`= 0 + 1/3 - (3x^2)/36`
`= 1/3 - x^2/12`
`("d"("AR"))/"dx" = 0`
`1/3 - x^2/12 = 0`
`1/3 = x^2/12`
`x^2 = 12/3`
x2 = 4
x = 2
`("d"^2 ("AR"))/("dx"^2) = 0 - (2x)/12 = - x/6`
When x = 2, `("d"^2 ("AR"))/("dx"^2) = - 2/6 = - 1/3`, negative
∴ AR is maximum when x = 2
Now, AR = `15 + x/3 - 1/36 x^3`
When x = 2, AR = `15 + 2/3 - 2^3/36`
`= 15 + 2/3 - 8/36`
`= 15 + (24 - 8)/36`
`= 15 + 16/36`
`= 15 + 4/9` ...(1)
R = 15x + `x^2/3 - 1/36 x^4`
Marginal Revenue (MR) = `"dR"/"dx"`
`= 15 + (2x)/3 - (4x)^3/36`
`= 15 + 2/3 x - x^3/9`
When x = 2, MR = 15 + `2/3 xx 2 - 2^3/9`
`= 15 + 4/3 - 8/9`
`= 15 + (12 - 8)/9`
`= 15 + 4/9` ....(2)
From (1) and (2) at the highest point average revenue is equal to the marginal revenue.
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