The demand for monthly cell phone plans is given by:

\[q_D = 50-0.5p\]

1. Write the inverse demand function.

Simply take the demand function and solve it for \(p\):

\[\begin{align*} q_D&=50-0.5p\\ q_D+0.5p&=50\\ 0.5p&=50-q_D\\ p&=100-2q_D\\ \end{align*}\]

The vertical intercept (“choke price”) is $100 and the slope is \(-2\).

2. Calculate the price elasticity of demand when the price is $10. Is this relatively elastic or inelastic?

First we need to find \(q_D\) at $10.

\[\begin{align*} q_D&=50-0.5(10)\\ q_D&=50-5\\ q_D&=45\\ \end{align*}\]

Now we have the three ingredients to calculate elasticity at $10:

\[\begin{align*} \epsilon_D &=\frac{1}{slope} \times \frac{p}{q_D} \\ \epsilon_D &=\frac{1}{-2} \times \frac{10}{45} \\ \epsilon_D &=-0.5 \times 0.22\\ \epsilon_D &=-0.11\\ \end{align*}\]

The demand is relatively inelastic, as \(|\epsilon_D|<1\)

3. Calculate the price elasticity of demand when the price is $70. Is this relatively elastic or inelastic?

First we need to find \(q_D\) at $70.

\[\begin{align*} q_D&=50-0.5(70)\\ q_D&=50-35\\ q_D&=15\\ \end{align*}\]

We already have the slope (since the demand is a straight line), so now we can simply plug into the elasticity formula:

\[\begin{align*} \epsilon_D &=\frac{1}{slope} \times \frac{p}{q_D} \\ \epsilon_D &=\frac{1}{-2} \times \frac{70}{15}\\ \epsilon_D &=-0.5 \times 4.67\\ \epsilon_D &\approx -2.33\\ \end{align*}\]

The demand is relatively elastic, as \(|\epsilon_D|>1\)

4. At what price is demand unit elastic \((\epsilon=-1)\)?

\[\begin{align*} \epsilon_D &=\frac{1}{slope} \times \frac{p}{q_D} & & \text{Formula for elasticity}\\ -1&=-0.5 \times \frac{p}{q_D} & & \text{Set } \epsilon_D \text{ equal to }-1\\ -1&=-0.5 \times \frac{p}{(50-0.5p)} & & \text{Plug in demand function for } q_D\\ -1(50-0.5p)&=-0.5p & & \text{Multiply by term in parentheses}\\ 0.5p-50&=-0.5p & & \text{Distribute the }-1 \\ -50&=-p & & \text{Add } 0.5p\\ p&=\$50 & & \text{Divide by }-50\\ \end{align*}\]

5. Calculate the total revenue at $10.

The total revenue is:

\[\begin{align*} R&=pq\\ R&=(\$10)(45)\\ R&=\$450\\ \end{align*}\]

6. Calculate the total revenue at $70.

The total revenue is:

\[\begin{align*} R&=pq\\ R&=(\$70)(15)\\ R&=\$1,050\\ \end{align*}\]

7. Calculate the total revenue at the price you found for question 4.

That price was \(p=\$50\). At this price, we need to find the quantity demanded. We can use the demand function:

\[\begin{align*} q_D&=50-0.5p\\ q_D&=50-0.5(50)\\ q_D&=50-25\\ q_D&=25\\ \end{align*}\]

Now that we have price and quantity, revenue is:

\[\begin{align*} R& = pq\\ R&=(\$50)(25)\\ R&=\$1,250\\ \end{align*}\]

This is where revenue is maximized.