Question

a. A manufacturer’s marginal revenue function is MR=105-x-0.3x^2. Find the increase in
the manufacturer’s total revenue if production is increased from 10 to 20 units.

b. A firm has a marginal revenue given by MR=(3/2x+7)-(1/20), where x is the output. Find
the corresponding demand function.

c. Given that the elasticity of demand for a commodity is given by e_xp=3-2p, where p denotes the price per unit of the commodity. find the demand function x.

Answer 1

a) $MR = 105 - x - 0.3x^2$

Total Revenue (TR) is found by integrating MR

$TR = \int MR dx$

$= \int (105 - x - 0.3x^2)dx$

$= [105x - \dfrac {x^2} {2} - 0.1x^3 + k]$

Increase in Total revenue is found by TR(20) - TR(10)

$TR(20) = 105(20) - \dfrac {20^2} {2} - 0.1(20)^3 + k$

$= 2100 - 200 - 800 + k$

$= \$(1,100 + k)$

$TR(10) = 105(10) - \dfrac {10^2} {2} - 0.1(10^3) + k$

$= 1050 - 50 - 100 + k$

$= \$(900 + k)$

Thus,

Increase in TR = $\$(1100 + k) - \$(900 + k)$

$= 1100 - 900 + k - k$

$= \$200$

b) $MR = \dfrac {3} {2x + 7} - \dfrac {1} {20}$

$TR = \int MR dx$

$= \int ( \dfrac {3} {2x + 7} - \dfrac {1} {20}) dx$

$= \dfrac {3} {2} ln|2x + 7| - \dfrac {1} {20} x + A$

$AR = \dfrac {TR} {x}$

$= \dfrac {\dfrac {3} {2} ln|2x + 7|} {x} - \dfrac {\dfrac {1} {20} x} {x} + \dfrac {A} {x}$

But, AR = Price (P)

Therefore, the demand function is given by:

$P = = \dfrac {3} {2x} ln|2x + 7| + \dfrac {A} {x} - \dfrac {1} {20}$ ,

where A is a constant.

c) $e_{xp} = 3 - 2p$

$\dfrac {dX} {dP} × \dfrac {p} {x} = 3 - 2p$

$\dfrac {dX} {x} = (\dfrac {3} {p} - 2)dP$

$=> \int (\dfrac {1} {x})dX = \int (\dfrac {3} {p} - 2)dP$

$=> ln|x| = 3ln|p| - 2p + lnA$

Where, lnA is a constant

$=> e^{lnx} = e^{lnp^3 - 2p + lnA}$

Therefore, the demand function is given by :

$x = Ap^3e^{-2p}$