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

MATH101 C April 2024

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[hide] Question 18 (c)
Suppose that over the entire market, ${\tilde {q}}_{e}$ units are sold, each at the price $S({\tilde {q}}_{e})$. Find the total surplus. You do not need to include units in your answer.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

[show] Hint

A formula for the total surplus when ${\tilde {q}}_{e}$ units are sold over the entire market is $${\begin{aligned}\int _{0}^{{\tilde {q}}_{e}}D(q)-S(q)\;\mathrm {d} q.\end{aligned}}$$ To intuitively remember this formula, and to understand why it is true, recall that each additional unit sold provides a surplus because the producer sells it for more than they are willing to sell it for, and the customer buys it for less than they are willing to spend it on, until the equilibrium is reached. This integral is the result of adding up those surpluses over all the units sold.

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[show] Solution

Whoops, no solution has been written yet. Be the first to submit a suggestion. 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 $${\begin{aligned}{\frac {1-2^{-15}}{\log 2}}+2\cdot 15-{\frac {15^{2}}{20}}-5={\frac {1-2^{-15}}{\log 2}}+30-{\frac {45}{4}}-5={\frac {1-2^{-15}}{\log 2}}+25-{\frac {45}{4}}.\end{aligned}}$$