In a certain chemical industry, the marginal private cost (MPC) of producing $Q$ thousand tonnes of product is given by $M P C = 40 + 3 Q$ (Rs per tonne). The marginal external damage (MED) from pollution is $M E D = 2 Q$ (Rs per tonne). The market price of the product is fixed at Rs.200 per tonne.

hard 4 marks Class 12 · Economics · Environment and Sustainable Development

Answer

Q_{e} = 53.33

thousand tonnes; b)

t^{*} = R s .64

per tonne; c)

Q_{t} = 32

thousand tonnes; d) Reduction in TSC = Rs.9,954 thousand

Step-by-step solution

Set MPC equal to market price to find unregulated equilibrium: $40 + 3 Q_{e} = 200$ .
Solve for $Q_{e}$ : $3 Q_{e} = 160 \Rightarrow Q_{e} = \frac{160}{3} = 53.33$ (thousand tonnes).
Social marginal cost (SMC) is $M P C + M E D = (40 + 3 Q) + 2 Q = 40 + 5 Q$ .
For optimal output, set SMC equal to market price: $40 + 5 Q^{*} = 200$ .
Solve for $Q^{*}$ : $5 Q^{*} = 160 \Rightarrow Q^{*} = 32$ (thousand tonnes).
Pigouvian tax $t^{*}$ equals the marginal external damage at the optimal output: $t^{*} = M E D_{Q = 32} = 2 \times 32 = 64$ Rs per tonne.
With the tax, firms face effective marginal cost $M P C + t^{*}$ ; set this equal to price: $40 + 3 Q_{t} + 64 = 200$ .
Solve for $Q_{t}$ : $3 Q_{t} = 96 \Rightarrow Q_{t} = 32$ thousand tonnes (matches $Q^{*}$ ).
Total social cost before tax: $T S C_{b e f or e} = (M P C + M E D) \times Q_{e} = (40 + 5 Q_{e}) \times Q_{e}$ .
Substitute $Q_{e} = 53.33$ : $T S C_{b e f or e} = (40 + 5 \times 53.33) \times 53.33 = 306.67 \times 53.33 \approx 16, 354.7$ thousand Rs.
Total social cost after tax (socially optimal): $T S C_{a f t er} = p r i ce \times Q_{t} = 200 \times 32 = 6, 400$ thousand Rs.
Reduction in total social cost = $T S C_{b e f or e} - T S C_{a f t er} \approx 16, 354.7 - 6, 400 = 9, 954.7$ thousand Rs (rounded to Rs.9,954 thousand).

Common mistake: Using the external damage at the unregulated output instead of at the optimal output to calculate the tax

Concept tested: Pigouvian tax and optimal output