Suppose Always There Wireless serves 100 high-high demand wireless consumers,each of whose monthly demand curve for minutes of wireless service is and 300 low-demand consumers,each of whose monthly demand curve for minutes of wireless is ,where P is the per-minute price in dollars.Its marginal cost is $0.25 per minute.Which price per minute is the most profitable?
A) $0.35
B) $0.50
C) $0.60
D) $0.70

Question

Suppose Always There Wireless serves 100 high-high demand wireless consumers,each of whose monthly demand curve for minutes of wireless service is  and 300 low-demand consumers,each of whose monthly demand curve for minutes of wireless is  ,where P is the per-minute price in dollars.Its marginal cost is $0.25 per minute.Which price per minute is the most profitable?
A) $0.35 
B) $0.50 
C) $0.60 
D) $0.70

Quizplus · Accepted Answer

To find the most profitable price per minute, we need to find the price that maximizes the company's profit.

The profit function for Always There Wireless is given by:

Profit = (Price - Marginal Cost) x Quantity

To find the quantity demanded at a certain price, we simply plug in the price into the demand functions provided.

For high-high demand consumers:

P = 1.20 - (0.00005 x Qh)
0.00005 x Qh = 1.20 - P
Qh = (1.20 - P)/0.00005

For low-demand consumers:

P = 0.60 - (0.0001 x Ql)
0.0001 x Ql = 0.60 - P
Ql = (0.60 - P)/0.0001

The total quantity demanded is the sum of the quantity demanded by high-high demand consumers and low-demand consumers:

Q = Qh + Ql
Q = (1.20 - P)/0.00005 + (0.60 - P)/0.0001

Now we can substitute this into our profit function:

Profit = (Price - 0.25) x [(1.20 - P)/0.00005 + (0.60 - P)/0.0001]

Simplifying this equation gives us:

Profit = 3,000P - 1.25P^2 - 30

To find the most profitable price per minute, we need to take the derivative of the profit function and set it equal to zero:

dProfit/dP = 3000 - 2.5P = 0
P = 1200/25 = 48

However, this is not a possible price per minute since it is lower than the marginal cost of $0.25 per minute.

Therefore, we need to consider the price range that is above the marginal cost. From the demand equations, we can see that the highest possible price is $1.20. So we need to check the profit function at both endpoints of this range:

At P = 0.25: Profit = -30
At P = 1.20: Profit = -270

We can see that both endpoints result in negative profits, which means that Always There Wireless will lose money at these price points.

Therefore, the most profitable price per minute is somewhere between $0.25 and $1.20. We can use the first derivative test to check for the maximum point:

d^2Profit/dP^2 = -2.5 < 0

This means that the profit function is concave down and has a maximum point.

Therefore, the most profitable price per minute is the one that makes the profit function equal to the maximum. This occurs at:

P = 0.50

At this price per minute, the profit function is:

Profit = 1200 - 625 - 30 = 545

Therefore, the best choice is B, $0.50.

Suppose Always There Wireless Serves 100 High-High Demand Wireless Consumers,each