1a)In this exercise B(x) is the marginal benefit of the firm while D(x) symbolizes the marginal
damage. We will denote with TB(x) and TD(x) the total benefits of the firm and the total
damage of the environment respectively. Then the social welfare function can be written as
the sum of the total benefit minus total damage. More specific:
W(x) = TB(x) – TD(x) . The welfare has to be maximised; thus the first order derivative of W
must be equal to zero: W’(x) = B(x) – D(x) = 0→ B(x) = D(x) .They are both given, so we
substitute, a – bx = d + gx => (b+g) x = a – d => x = (a – d)/(b + g).
Exercise 1 b)
Coase Theorem
In a world in which some producers or consumers are subject to externalities, generated by
other consumers or producers, assume:
1)
2)
3)
4)
5)
6)
Perfect information
Everyone are price takers
There is a costless force system for enforcing agreements
Producers maximize profits and consumers maximize utility
There are no income or wealth effects
No transaction costs
In this case, the initial assignment of property rights regarding externalities does not matter
for efficiency.
If any of these assumptions are violated, the initial assignment of the rights does matter.
For our problem this implies that efficiency will be achieved whether the polluter has a right
to pollute, or the consumer has a right to have a clean environment. The initial assignment of
rights is a purely a distributional issue. But does the initial distribution of right to pollute
matters for social optimum?
The most controversial assumption is probably no transaction costs, which cover different
costs, associated with agreeing to transact.
Lets saythat c is a transaction costs, which means that the consumer have to pay additinal cost
c perreduced unit of pollution, so the Marginal damage is D(x)= d - c + gx. This gives us
higher level of pollution then it should be.Moreover, if c exeeds payoff of transaction, the
transaction would not take place at all.
1c) Suppose the regulator wants to apply an emission tax t, such that will determine the
optimal emissions on x*. Assume furthermore that without any intervention, the firm i was
planning to produce until MS(x0) = 0 in point x0. From the reduction in emissions equal to x0x* that the fee will cause, the firm will have an additional cost equal to t*x*. So without any
side payment it will not be willing to accept the tax.
If the emissions reduce to x*, two things happen: there is a benefit to the consumers equal
to ∫x*xo(D(x)) dx . Furthermore, there is a loss in benefits for the company equal to
∫x*xo(B(x))dx. Handling with these two definite integrals is manageable since we know that the
indefinite integrals of D(x) and B(x) are the total damage of x TD(x) and the total benefits of
x TB(x) respectively. So the net social gain from the tax application is :
Net social gain:

∫x*xo(D(x)) dx -∫x*xo(B(x))dx = [TD(x)]xox* - [TB(x)] xox* =
=TD(x0) – TD(x*)-TB(x0) +TB(x*). This is the yellow area. The green area is what the firm is
Called to pay. Now imagine that victims offer the entire yellow area as a lump sum transfer to
the company. If the yellow area is bigger than the green then the firm has an incentive to
accept the offer. Mathematically, TD(x0) – TD(x*)-TB(x0) +TB(x*) > t*x*. The yellow
surface depends on the slopes of B(x) and D(x). The steeper they are, the bigger it gets.
2a) The problem of the firm is profit maximization. The profit is π = pz – wx – ry =>
π = p x1/3y1/3 – wx – ry . So max p x1/3y1/3 – wx – ry. First order conditions are:
∂ π /∂x = 0 => 1/3 [px -2/3 y1/3] = w , and
∂ π /∂y = 0 => 1/3 [px 1/3y -2/3] = r . By taking the ratio of the two first order conditions we
end up to y*/x* = w/r. And that is the optimal proportion of the x* and y* use for profit
maximization.
We can also solve the dual problem, the cost minimization and check for solutions.
Min C(x,y) subject to z = x1/3y1/3. Lagrangian for that problem then is
ℓ= C(x,y) – λ (z - x1/3y1/3)= wx + ry – λ (z - x1/3y1/3). And the first order conditions are
∂ ℓ /∂x = 0 => w = - 1/3[λ x -2/3 y1/3] (1)
∂ ℓ /∂y = 0 => r = - 1/3[λx 1/3y -2/3] (2)
z = x1/3y1/3 (3)
Again (1)/(2) gives y*/x* = w/r => y* = x*[w/r]
2b)
2c)
Inserting the optimal value of x as a function of y, x*=y*[r/w], gives profit as a function of y,
r and w:
 (r , w, y)  p(r w)1/ 3 y 2/ 3  2ry