Chapter 6.7, Problem 6E

To determine

To Calculate: The producer surplus at a certain sales level for a given supply function.

Answer to Problem 6E

The producer surplus is $\$6.\overline{6}$

Explanation of Solution

Given Information: The supply function is $p_s(x)=3+0.01x^2$ and the sales level is $X=10$

Method Used: The producer surplus at a particular selling price $P$ and a supply function $p_s\left(x\right)$ is

  $ProducerSurplus=\underset{0}{\overset{X}{{\displaystyle \int }}}\left[P-p_s(x)\right]dx$

Where $P$ is the value of $p_s\left(X\right)$ , the price that is to be set to achieve $X$ level of sales.

Calculation: Put the value of $X=10$ in the supply function to write

  $P=p_s(10)=3+0.01\times {10}^2=4$

Now put $P,p_s\left(x\right)$ and $X$ in the integral for the producer surplus to get

  $ProducerSurplus=\underset{0}{\overset{10}{{\displaystyle \int }}}\left(4-[3+0.01x^2]\right)dx=\underset{0}{\overset{10}{{\displaystyle \int }}}(1-0.01x^2)dx$

  $={x-0.01\times \frac{x^3}{3}]}_0^{10}=10-\frac{0.01\times 1000}{3}$

  $=6.\overline{6}$

Therefore, the producer surplus is $\$6.\overline{6}$

Graph: The producer surplus is the area enclosed by thick black lines in the plot shown below.