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

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

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

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

The consumer surplus is:

${\begin{aligned}CS&=\int _{0}^{A}D(x)dx-Ap_{A}\\&=\int _{0}^{4}{\frac {42}{x+2}}dx-(4)(7)\\&=42\ln(x+2){\big |}_{0}^{4}-28\\&=42(\ln(6)-\ln(2))-28\\&=42\ln(3)-28\\&\approx 18.14\end{aligned}}$

The producer surplus is:

${\begin{aligned}PS&=Ap_{A}-\int _{0}^{A}S(x)dx\\&=(4)(7)-\int _{0}^{4}(x+3)dx\\&=28-(x^{2}/2+3x){\big |}_{0}^{4}\\&=28-(16/2+12)\\&=8\end{aligned}}$