Question

The demand function is $x =\dfrac{{24 - 2p}}{3}$ where x is the number of units demanded and p is the price per unit. Find:
    i)The revenue function R in terms of p.
    ii)The price and the number of units demanded for which the revenue is maximum.

Answer 1

Hint:
As per unit price p is given and x number of units are demanded, multiplying these we can find the revenue function R. Then we have to investigate the value of p and x for which R will be maximum. Then second order derivatives at that point will be negative.

Complete step by step solution:
Step1:i) Since, p is the price per unit and demand is x units,then the revenue will be
                       $\eqalign{
  & R = p \times x \cr
  & or,R(p) = p \times\dfrac{{(24 - 2p)}}{3} \cr
  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 8p -\dfrac{2}{3}{p^2} \cr} $.
Here R(p) means that R is a function of p.

Step2: ii)
  $$\eqalign{
  & R(p) = 8p -\dfrac{2}{3}{p^2} \cr
  & Differentiating\,\,\,w.r.t.\,\,\,p, \cr
  &\dfrac{{dR}}{{dp}} = 8 -\dfrac{4}{3}p \cr
  & Again\,\,differentiating\,\,w.r.t.\,\,\,\,p, \cr
  &\dfrac{{{d^2}R}}{{d{p^2}}} = 0 -\dfrac{4}{3} = -\dfrac{4}{3} \cr
  & Now,\dfrac{{dR}}{{dp}} = 0\,\,\,gives \cr
  & \,\,\,\,\,\,\,8 -\dfrac{4}{3}p = 0 \cr
  & or,\dfrac{4}{3}p = 8 \cr
  & or,p = 6 \cr
  & Now,\,\,at\,\,p = 6, \cr
  &\dfrac{{{d^2}R}}{{d{p^2}}} = -\dfrac{4}{3} < 0 \cr} $$
Therefore R has maximum value at p=6.
Again,
  $\eqalign{
  & x =\dfrac{{24 - 2p}}{3} \cr
  & \,\,\,\,\, =\dfrac{{24 - (2 \times 6)}}{3} \cr
  & \,\,\,\,\, =\dfrac{{12}}{3} \cr
  & \,\,\,\,\, = 4 \cr} $

Therefore the price and the number of units demanded are 6 and 4 respectively for which the revenue is maximum.

Note:
We have to check the sign of the second order derivative at the point where maximum or the minimum value may occur. For maximum the sign will be negative and for the minimum it will be positive.