Science:Math Exam Resources/Courses/MATH105/April 2011/Question 04/Solution 1

Let us write the supply and demand function as functions in the price $p$. The text says that at a price $p$ the supplier produces $p-2$ units. Thus, the supply function must be

$\displaystyle p=S(q)=q+2$

At a price $p$ the consumer demands $65/p-10$ units. In terms of the demand function this translates into

$p=D(q)={\frac {65}{q+10}}$

By definition, the market equilibrium is the intersection point of the graph of $S(q)$ and $D(q)$. Let us denote this point by $(q_{e},p_{e})$. To compute that point, we set both functions equal and compute a solution for $q$:

${\begin{aligned}S(q)&=D(q)\\q+2&={\frac {65}{q+10}}\end{aligned}}$

And rearranging we obtain the equation

${\begin{aligned}(q+2)(q+10)&=65\\q^{2}+12q-45&=0\end{aligned}}$

This is a quadratic equation and can be solved with the help of the $abc$-formula: the solutions to a quadratic equation of the form $ax^{2}+bx+c=0$ are given by

$x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}$

Note, that in our problem the variable $x$ is replaced by the variable $q$ and we have $a=1$, $b=12$ and $x=-45$. Hence,

$q={\frac {-12\pm {\sqrt {144+180}}}{2}}=-6\pm 9$

The solution $-6-9$ is not feasible, because quantities are always nonnegative. Hence, we arrive at the unique solution $q_{e}=-6+9=3$. We can calculate the associated price by using $D(q)$ or $S(q)$. Since $S(q)$ looks simpler than $D(q)$ we use $S(q)$ and get $p=S(q_{e})=S(3)=3+2=5$. Note, that this price is in between the range given in the problem. Let us summarize:

The market equilibrium is attained at $(q_{e},p_{e})=(3,5)$

The second part of the question asks us to compute the consumer surplus. By definition, the consumer surplus is given by the formula

${\text{consumer surplus}}=\int _{0}^{q_{e}}(D(q)-p_{e})\,dq$

The following steps compute this integral:

${\begin{aligned}\int _{0}^{q_{e}}(D(q)-p_{e})\,dq&=\int _{0}^{3}\left({\frac {65}{q+10}}-5\right)\,dq\\&=65\int _{0}^{3}{\frac {1}{q+10}}\,dq-\int _{0}^{3}5\,dq\\&=65\ln(q+10){\Bigr |}_{0}^{3}-5q{\Bigr |}_{0}^{3}\\&=65(\ln(13)-\ln(10))-15\\&=65\ln(13/10)-15\\&\approx 2.05\end{aligned}}$

The consumer surplus is approximately $2.05