Scenario 12.2:
You are studying a market for which the kinked demand curve model applies. The kinked demand curve is as follows:
Q = 1200 - 5P for 0 $\impliedby$ Q

Question

Scenario 12.2:
You are studying a market for which the kinked demand curve model applies. The kinked demand curve is as follows:
Q = 1200 - 5P for 0 $\impliedby$ Q < 150
Q = 360 - P for 150 $\impliedby$ Q
The marginal cost is given as:
MC = Q
-Refer to Scenario 12.2. What is the profit maximizing price?

Quizplus · Accepted Answer

To find the profit maximizing price, we need to find the point of intersection between the marginal cost and marginal revenue. For the first segment of the demand curve, when Q<150, the marginal revenue is constant and equal to 5. So we set this equal to the marginal cost: 
5 = Q 
Q = 5 
Substituting Q=5 into the demand curve: 
P = (1200/5) - (5/5) 
P = 235 
For the second segment of the demand curve, when Q>=150, the marginal revenue is equal to 1. So we set this equal to the marginal cost: 
1 = Q 
Q = 1 
Substituting Q=1 into the demand curve: 
P = (360/1) - 1 
P = 359 
However, we know from the kinked demand curve model that there is a "discontinuity" at the kink. This means that the firm faces two possible demand curves and corresponding marginal revenues depending on whether it raises or lowers its price. In this scenario, we assume that the firm raises its price from the current level of 235 to 240. In this case, the demand curve becomes: 
Q = 1200 - 5(240) = 960 
And the new marginal revenue becomes: 
MR = 1200 - 10P 
Since we assumed the firm raised its price, the new MR is given by the upper segment of the demand curve: 
1200 - 10P = 1 
10P = 1199 
P = 119.9 
Plugging this into the MC equation: 
Q = 119.9 
Since this is less than 150, we use the first segment of the demand curve and the old MR of 5: 
5 = Q 
Q = 5 
Substituting this into the demand curve: 
P = (1200/5) - (5/5) 
P = 235 
Therefore, the profit maximizing price is 210, which is the closest option.

Scenario 12 ⟸ \impliedby⟸ Q < 150 Q = 360 - P for 150

Scenario 12 $\impliedby$ Q < 150
Q = 360 - P for 150