Science:Math Exam Resources/Courses/MATH101 C/April 2024/Question 18 (c)/Solution 1

Let us keep the variable ${\tilde {q}}_{e}$ general to start, so that we can answer the question for both ${\tilde {q}}_{e}=20$ (the true value) and ${\tilde {q}}_{e}=15$ (the value the test said to use if you did not get part (b)). (Note that both $15$ and $20$ are greater than $10$, so we can assume ${\tilde {q}}_{e}>10$ in our calculations.)

Using the formula mentioned in the hint, the surplus is

$${\begin{aligned}\int _{0}^{{\tilde {q}}_{e}}D(q)-S(q)&=\int _{0}^{10}\left({\frac {1}{2^{q}}}+1\right)\;\mathrm {d} q+\int _{10}^{{\tilde {q}}_{e}}\left({\frac {1}{2^{q}}}+1-\left({\frac {q}{10}}-1\right)\right)\;\mathrm {d} q\\&=\int _{0}^{10}\left(2^{-q}+1\right)\;\mathrm {d} q+\int _{10}^{{\tilde {q}}_{e}}\left(2^{-q}+2-{\frac {q}{10}}\right)\;\mathrm {d} q.\end{aligned}}$$

Note that $\int 2^{-q}\;\mathrm {d} q=\int e^{-(\log 2)q}\;\mathrm {d} q={\frac {1}{\log 2}}(-e^{-(\log 2)q})+C=-{\frac {1}{\log 2}}2^{-q}+C$ (where $\log$ is the natural logarithm), so the surplus is equal to

$${\begin{aligned}&\left[-{\frac {1}{\log 2}}2^{-q}+q\right]{\Bigg |}_{0}^{10}+\left[-{\frac {1}{\log 2}}2^{-q}+2q-{\frac {q^{2}}{20}}\right]{\Bigg |}_{10}^{{\tilde {q}}_{e}}\\=&\left(-{\frac {1}{\log 2}}2^{-10}+10\right)-\left(-{\frac {1}{\log 2}}\right)+\left(-{\frac {1}{\log 2}}2^{-{\tilde {q}}_{e}}+2{\tilde {q}}_{e}-{\frac {{\tilde {q}}_{e}^{2}}{20}}\right)-\left(-{\frac {1}{\log 2}}2^{-10}+20-{\frac {100}{20}}\right)\\=&{\frac {1}{\log 20}}(-2^{-10}+1-2^{-{\tilde {q}}_{e}}+2^{-10})+10+2{\tilde {q}}_{e}-{\frac {{\tilde {q}}_{e}^{2}}{20}}-20+5\\=&{\frac {1-2^{-{\tilde {q}}_{e}}}{\log 2}}+2{\tilde {q}}_{e}-{\frac {{\tilde {q}}_{e}^{2}}{20}}-5.\end{aligned}}$$

Plug in ${\tilde {q}}_{e}=20$. The surplus is

$${\begin{aligned}{\frac {1-2^{-20}}{\log 2}}+2\cdot 20-{\frac {20^{2}}{20}}-5={\frac {1-2^{-20}}{\log 2}}+40-20-5={\frac {1-2^{-20}}{\log 2}}+15.\end{aligned}}$$

For people who plugged in $15$ instead of $20$, the answer should instead be

$${\begin{aligned}{\frac {1-2^{-15}}{\log 2}}+2\cdot 15-{\frac {15^{2}}{20}}-5={\frac {1-2^{-15}}{\log 2}}+30-{\frac {45}{4}}-5={\frac {1-2^{-15}}{\log 2}}+25-{\frac {45}{4}}.\end{aligned}}$$