Question: The demand function for a Christmas music CD is given by
$x = 0.25(225 - p^2)$
where $x$ (measured in units of a hundred) is the quantity demanded per week and $p$ is the unit price in dollars.
(a) Evaluate the elasticity at 10. $E(10) =$
(b) Should the unit price be lowered slightly from 10 in order to increase revenue?
(c) When is the demand unit elastic? $p = $ dollars
(d) Find the maximum revenue. Maximum revenue = hundreds of dollars
1. **State the problem:**
We have the demand function $$x = 0.25(225 - p^2)$$ where $x$ is quantity demanded (in hundreds) and $p$ is price.
2. **Find the elasticity formula:**
Elasticity of demand is $$E(p) = \frac{dx}{dp} \cdot \frac{p}{x}$$
3. **Calculate $\frac{dx}{dp}$:**
$$x = 0.25(225 - p^2) = 56.25 - 0.25p^2$$
$$\frac{dx}{dp} = -0.5p$$
4. **Evaluate elasticity at $p=10$:**
Calculate $x$ at $p=10$:
$$x = 0.25(225 - 10^2) = 0.25(225 - 100) = 0.25 \times 125 = 31.25$$
Calculate $E(10)$:
$$E(10) = (-0.5 \times 10) \times \frac{10}{31.25} = (-5) \times \frac{10}{31.25} = -\frac{50}{31.25} = -1.6$$
5. **Interpret elasticity at $p=10$:**
Since $|E(10)| = 1.6 > 1$, demand is elastic, so lowering price will increase revenue.
6. **Find when demand is unit elastic:**
Set $|E(p)| = 1$:
$$|E(p)| = \left| -0.5p \times \frac{p}{0.25(225 - p^2)} \right| = 1$$
Simplify:
$$0.5p \times \frac{p}{0.25(225 - p^2)} = 1$$
$$\frac{0.5p^2}{0.25(225 - p^2)} = 1$$
$$\frac{0.5p^2}{56.25 - 0.25p^2} = 1$$
Multiply both sides:
$$0.5p^2 = 56.25 - 0.25p^2$$
$$0.5p^2 + 0.25p^2 = 56.25$$
$$0.75p^2 = 56.25$$
$$p^2 = \frac{56.25}{0.75} = 75$$
$$p = \sqrt{75} = 5\sqrt{3} \approx 8.66$$
7. **Find maximum revenue:**
Revenue $R = p \times x = p \times 0.25(225 - p^2) = 0.25p(225 - p^2) = 56.25p - 0.25p^3$
Find critical points by setting derivative to zero:
$$\frac{dR}{dp} = 56.25 - 0.75p^2 = 0$$
$$0.75p^2 = 56.25$$
$$p^2 = 75$$
$$p = 5\sqrt{3} \approx 8.66$$
Calculate maximum revenue:
$$R(8.66) = 0.25 \times 8.66 \times (225 - 8.66^2) = 0.25 \times 8.66 \times (225 - 75) = 0.25 \times 8.66 \times 150 = 324.75$$
Since $x$ is in hundreds and $R$ is in hundreds of dollars, maximum revenue is approximately 325 hundreds of dollars.
**Final answers:**
(a) $E(10) = -1.6$
(b) Yes, lowering price from 10 will increase revenue.
(c) Demand is unit elastic at $p = 5\sqrt{3} \approx 8.66$
(d) Maximum revenue is approximately 325 hundreds of dollars.