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

$$Solution$$

a.) Marginal revenue function is $MR=105-x-0.3x^2$ . Production is increased from 10 to 20 units.

Note that in calculus terms, the marginal revenue (MR) is the first derivative of the total revenue (TR) function with respect to the quantity:

$$MR = \frac{ΔTR}{ΔQ}$$

To calculate the total revenue, we need to integrate the marginal revenue function

$$TR=\intop^{20}_{10} (105-x-0.3x^2)dx$$

$\implies\int _{10}^{20}105dx-\int _{10}^{20}xdx-\int _{10}^{20}0.3x^2dx=200.000001$

Total revenue is $=200.000001-------->Answer$

b.) $MR=\frac{3}{2x}+7-\frac{1}{20}$ , where $x$ is the output.

The price per unit $p$ is also called the demand function $p$

$Revenue = (price per unit) . (number of units)$

$R(x)=\intop R'(x)=\int (\frac{3}{2x}+7-\frac{1}{20})dx=\frac{3}{2}ln|x|+\frac{139}{20}x+c$

To find the price per unit:

$p=\frac{R(x)}{x}=\frac{\frac{3}{2}ln|x|+\frac{139}{20}x+c}{x}$

$p=\frac{30ln|x|+139x+C.20}{20x}$ $------------->Answer$

c.) e_xp=3-2p, where p denotes the price per unit of the commodity. find the demand function x.

$\frac{dX}{dp}​*\frac{p}{x}​=3−2p$

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

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

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

Where lnA is a constant

$\implies \epsilon^{lnx}=\epsilon^{lnp^3-2p-lnA}$

$AP^3\epsilon ^{-2p}$ $-------->Answer$