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Oscar’s demand for movies is given by Q = 10−2P.

(a) What is the price elasticity of demand at a price of 2? Is Oscar’s demand elastic or inelastic at a price of 2?

(b) Assume that the price is 3. What is Oscar’s total expenditure on movies? What is the consumer surplus?

(c) What is the price that maximizes Oscar’s total expenditure on movies? What is the price elasticity of demand at this price?

(d) If the price increases from 1 to 2, does Oscar’s total expenditure on movies rise or fall? If the price rises from 3 to 4, does Oscar’s total expenditure on movies rise or fall?

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closed as off-topic by Giskard, Herr K., EconJohn Sep 14 '17 at 16:34

This question appears to be off-topic. The users who voted to close gave this specific reason:

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(a) Price elasticity is $\eta_d=\frac{\partial Q}{\partial P} \cdot \frac{P}{Q}$. Therefore you get $\eta_d=-2 \cdot \frac{2}{10-2\cdot2}= -\frac{2}{3}$.

(b) $Q=10-2 \cdot 3 = 4$. Total expenditure is $3 \cdot 4 = 12$. At a price of 5 units, there is no demand. The consumer surplus for linear demand functions follows $\frac{1}{2}\left(p_{max}-p\right)\cdot x$ with $p_{max}$ as the price where there is no demand. Consumer surplus therefore is $\frac{1}{2}\left(5-3\right)\cdot 4=4$.

(c) Total expenditure $TR$ follows $TR=p \cdot x$. For maximizing $TR$ just derive it (and set it to zero) and you get $\frac{\partial TR}{\partial p}=\frac{\partial 10p-2p^2}{\partial p}=10-4\cdot p = 0$. This gives the value $2.5$ for $p$. On how to calculate elasticity see (a).

(d) Total expenditure for $p=1$ is $8 \cdot 1 = 8$, for $p=2$ it is $6 \cdot 2 = 12$. If price increases from 1 to 2, $TR$ increases. Look the result in (c), that $p=2.5$ maximizes $TR$. Therefore from $p=1$ to $p=2$, $TR$ has to increase. In equivalent, $TR$ has to decrease when price rises from $p=3$ to $p=4$.

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  • 3
    $\begingroup$ Should I? $\endgroup$ – Giskard Sep 14 '17 at 15:08
  • $\begingroup$ @denesp what is clear is that der_bosmann should have not. $\endgroup$ – luchonacho Sep 14 '17 at 18:11
  • $\begingroup$ Thanks for your helpful comments. As i am pretty new to the SE community, i try my best to contribute to this platform. I will keep your comments in mind for the future, but please. $\endgroup$ – skoestlmeier Sep 15 '17 at 9:22

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