Question: Currently, 10,000 people take city buses each day and pay 2 dollars for a ticket. The number of people taking city buses at price $p$ dollars per ticket is given $x = 5000\sqrt{6 - p}$ (a) Evaluate the elasticity at 2. $E(2) =$ (b) Currently, the price per ticket is 2 dollars. Should the price be raised in order to increase revenue? Enter 1 for Yes or 2 for No. answer = (c) For what value of $p$ is the demand is unit elastic? $p =$ (d) What is the maximum revenue? Revenue =

1. **State the problem:** We have a demand function $x = 5000\sqrt{6 - p}$ where $x$ is the number of people taking the bus and $p$ is the price per ticket. 2. **Elasticity formula:** Elasticity of demand $E(p)$ is given by $$E(p) = \frac{dx}{dp} \cdot \frac{p}{x}$$ 3. **Find $\frac{dx}{dp}$:** $$x = 5000(6 - p)^{1/2}$$ Using the chain rule, $$\frac{dx}{dp} = 5000 \cdot \frac{1}{2}(6 - p)^{-1/2} \cdot (-1) = -\frac{2500}{\sqrt{6 - p}}$$ 4. **Evaluate elasticity at $p=2$:** Calculate $x$ at $p=2$: $$x = 5000 \sqrt{6 - 2} = 5000 \sqrt{4} = 5000 \times 2 = 10000$$ Calculate $\frac{dx}{dp}$ at $p=2$: $$\frac{dx}{dp} = -\frac{2500}{\sqrt{6 - 2}} = -\frac{2500}{2} = -1250$$ Now, $$E(2) = (-1250) \times \frac{2}{10000} = -\frac{2500}{10000} = -0.25$$ 5. **Interpret elasticity:** Since $|E(2)| = 0.25 < 1$, demand is inelastic at $p=2$. 6. **Should price be raised to increase revenue?** When demand is inelastic, increasing price increases revenue. So answer = 1 (Yes). 7. **Find $p$ for unit elasticity:** Set $|E(p)| = 1$: $$|E(p)| = \left| \frac{dx}{dp} \cdot \frac{p}{x} \right| = 1$$ Substitute $\frac{dx}{dp} = -\frac{2500}{\sqrt{6 - p}}$ and $x = 5000 \sqrt{6 - p}$: $$1 = \left| -\frac{2500}{\sqrt{6 - p}} \cdot \frac{p}{5000 \sqrt{6 - p}} \right| = \frac{2500 p}{5000 (6 - p)} = \frac{p}{2(6 - p)}$$ Solve: $$1 = \frac{p}{2(6 - p)} \Rightarrow 2(6 - p) = p \Rightarrow 12 - 2p = p \Rightarrow 12 = 3p \Rightarrow p = 4$$ 8. **Maximum revenue:** Revenue $R = p \times x = p \times 5000 \sqrt{6 - p}$ To maximize $R$, differentiate and set to zero: $$R = 5000 p (6 - p)^{1/2}$$ $$\frac{dR}{dp} = 5000 \left( (6 - p)^{1/2} + p \cdot \frac{1}{2} (6 - p)^{-1/2} (-1) \right) = 5000 \left( \sqrt{6 - p} - \frac{p}{2 \sqrt{6 - p}} \right)$$ Set $\frac{dR}{dp} = 0$: $$\sqrt{6 - p} = \frac{p}{2 \sqrt{6 - p}} \Rightarrow (\sqrt{6 - p})^2 = \frac{p}{2}$$ $$6 - p = \frac{p}{2} \Rightarrow 6 = \frac{3p}{2} \Rightarrow p = 4$$ Calculate max revenue: $$R_{max} = 4 \times 5000 \sqrt{6 - 4} = 4 \times 5000 \times \sqrt{2} = 20000 \sqrt{2} \approx 28284.27$$ **Final answers:** (a) $E(2) = -0.25$ (b) answer = 1 (c) $p = 4$ (d) Revenue $= 20000 \sqrt{2} \approx 28284.27$