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.

Answer 1

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.