Question 11.10: Assume that there are two sources of pollution into a lake. ......
Assume that there are two sources of pollution into a lake. The local water authority can clean up the discharges and reduce pollution levels from these sources but there are, of course, costs involved. The damage effects of each pollution source are measured on a ‘pollution scale’. The lower the pollution level the greater the cost of achieving it, as is shown by the cost schedules for cleaning up the two pollution sources:
Z_1 = 478 − 2C_1^{0.5} and Z_2 = 600 − 3C_2^{0.5}
where Z_1 and Z_2 are pollution levels and C_1 and C_2 are expenditure levels (in £000s) on reducing pollution.
To secure an acceptable level of water purity in the lake the water authority’s objective is to reduce the total pollution level to 1,000 by the cheapest method. How can it do this?
This can be formulated as a constrained optimization problem where the constraint is the total amount of pollution Z_1 + Z_2 = 1000 and the objective function to be minimized is the cost of pollution control TC = C_1 + C_2 . Thus the Lagrange function is
G = C_1 + C_2 + λ(1,000 − Z_1 − Z_2)
Substituting in the cost functions for Z_1 and Z_2 , this becomes
G = C_1 + C_2 + λ[1,000 − (478 − 2C_1^{0.5} ) − (600 − 3C_2^{0.5} )]
G = C_1 + C_2 + λ(−78 + 2C_1^{0.5} + 3C_2^{0.5} )
First-order conditions require
\frac{\partial G}{\partial C_1 } = 1 + λC_1^{-0.5} = 0 giving λ = – C_1^{0.5} (1)
\frac{\partial G}{\partial C_2 } = 1 + λ1.5C_2^{- 0.5} = 0 giving λ = \frac{- C_2^{0.5}}{1.5} (2)
\frac{\partial G}{\partial \lambda } = −78 + 2C_1^{0.5} + 3C_2^{0.5} = 0 (3)
Equating (1) and (2)
– C_1^{0.5} = \frac{- C_2^{0.5}}{1.5}
1.5C_1^{0.5} = C_2^{0.5} (4)
Substituting (4) into (3)
−78 + 2C_1^{0.5} + 3(1.5C_1^{0.5} ) = 0
2C_1^{0.5} + 4.5C_1^{0.5} = 78
6.5C_1^{0.5} = 78
C_1^{0.5} = 12
C_1 = 144 (5)
Substituting (5) into (4)
C_2^{0.5} = 1.5(12) = 18
C_2 = 324
We can use these optimum pollution control expenditure amounts to check the total pollution level:
Z_1 + Z_2 = [(478 − 2C_1^{0.5} ) + (600 − 3C_2^{0.5} )]
= 478 − 2(12) + 600 − 3(18)
= 1,000
which is the required level. Thus the water authority should spend £144,000 on reducing the first pollution source and £324,000 on reducing the second source