Science:Math Exam Resources/Courses/MATH105/April 2010/Question 02/Solution 1

First we find the sale level A, where D(A) = S(A):

$\begin{array}{rl}D(A)& =S(A)\\ \frac{42}{A+2}& =A+3\\ 42=A^2+5A+6\\ A^2+5A-36=0\\ (A+9)(A-4)=0\\ A=-9\text{ or }A=4\end{array}$

The sale level must be positive, so A = 4. At this sale level, the price is p_A = 7.

The consumer surplus is:

$\begin{array}{rl}CS& =\underset{0}{\overset{A}{\int }}D(x)dx-A{p}_A\\ & =\underset{0}{\overset{4}{\int }}\frac{42}{x+2}dx-(4)(7)\\ & =42\ln (x+2){|}_0^4-28\\ & =42(\ln (6)-\ln (2))-28\\ & =42\ln (3)-28\\ & \approx 18.14\end{array}$

The producer surplus is:

$\begin{array}{rl}PS& =A{p}_A-\underset{0}{\overset{A}{\int }}S(x)dx\\ & =(4)(7)-\underset{0}{\overset{4}{\int }}(x+3)dx\\ & =28-(x^2/2+3x){|}_0^4\\ & =28-(16/2+12)\\ & =8\end{array}$