Question

The marginal (external) damage cost from air pollutant emission is MD = 20 + Q where Q is the quantity units of pollutants emitted. The marginal control cost associated with pollution cleanup is MC = 200-5Q.

(a) Draw the marginal damage cost and the marginal control cost together.

(b) What is the optimal (or efficient) level of pollution emission? In other words, what is the optimal level of pollution reduction?

(c) Compute the total damage cost and the total control cost at the optimal level of emission.

(d) Suppose the regulator limits the emission at 20 units. Compute the total damage cost and total control cost at that level. Find the sum of the total damage cost and total control cost and compare it to that of optimal emission level. Is the regulator achieving cost-efficiency by limiting pollutants to 20 units?

Answer 1

Red: MC, Blue: MD

optimal level of pollution, MD=MC
=> 20 + Q = 200 - 5Q
=> 6Q =180
=> Q = 30
In the graph above also, Q=30, where the curves intersect

Total damage cost = integral over Q=0 to 30
= 20Q + Q^2/2
= 20*30 + 30*30/2
= 1050

Total control cost = integral over Q=0 to 30
= 200Q - 5Q^2/2
= 200*30 - 5*30*30/2
= 3750

Q=20

Total damage cost = integral over Q=0 to 20
= 20Q + Q^2/2
= 20*20 + 20*20/2
= 600

Total control cost = integral over Q=0 to 20
= 200Q - 5Q^2/2
= 200*20 - 5*20*20/2
= 3000

Total control + damage cost:
Q=30; 1050+3750 = 4800
Q=20, 600+3000 = 3600

Cost efficiency is cost per unit output, it is as follows:
Q=30; 4800/30 = 160
Q=20, 3600/20 = 180
Since the cost per unit output increases, 20 units does not achieve cost efficiency