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I am trying to solve the following question:

Let $h \geq 0$ represent a negative externality of a firm's production on one (representative) consumer. The consumer has a quasi-linear utility function and attaches a utility of $\phi(h) = -2h^2$ to the externality. The firm's profit function is $\pi(h)=120-2(h-10)^2 $. Suppose the consumer has the property right concerning $h$ and can sell the right to produce a quantity $h$ at some price $P$. The consumer's utility function from some combination $(h,P)$ is $u(h,P)= \phi(h) + P$. The firm's profit function is $\Pi(h,P)=\pi(h)-P$.

  1. Suppose the firm can make the consumer a take-it-or-leave-it offer for an externality level $h$ at price $P$. If the consumer rejects the firm's offer the firm does not produce so the consumer's utility is 0. Compute the firm's optimal offer $(h_p^f,P^f)$ to the consumer.
  2. Suppose the consumer can make the firm a take-it-or-leave-it offer for an externality level $h$ at price $P$. If the firm rejects the offer it cannot produce so the firm's profit is 0. Compute the consumer's optimal offer $(h_p^c,P^c)$ to the firm.

Edit: Approach using the hint from Herr K.

  1. The consumer can reject the offer giving him a guaranteed utility of 0. Maximize the firm's profit subject to this constraint, $u(h,P)= \phi(h) + P = 0$. Solve this using Lagrangian method. $$ Lagrangian = objective function + constraint $$ $$\mathcal{L}(h,P,\lambda) = \Pi(h,P) + \lambda(\phi(h)+P)$$ $$\mathcal{L}(h,P,\lambda) = \pi(h) - P + \lambda(-2h^2+P)$$ $$\mathcal{L}(h,P,\lambda) = 120-2(h-10)^2 - P + \lambda(-2h^2+P)$$ $$\frac{\partial\mathcal{L}}{\partial h} = -4(h-10)-4h\lambda = 0 \rightarrow \lambda = \frac{-4h+40}{4h}$$ $$\frac{\partial\mathcal{L}}{\partial P} = -1 + \lambda = 0 \rightarrow \lambda = 1$$ $$\frac{\partial\mathcal{L}}{\partial \lambda} = -2h^2+P = 0$$

When we set both $\lambda$ equal to each other we get $1=\frac{-4h+40}{4h} \rightarrow h=5$. Plugging this into $\frac{\partial\mathcal{L}}{\partial \lambda}$ we get $-2(5)^2+P=0 \rightarrow 50=P$

Thus $h_p^f = 5$ and $P^f= 50$

  1. We do the same thing but this time the objective function is the consumer's utility function $u(h,P)= \phi(h) + P$ and the constraint is the firm's profit function $\Pi(h,P)=\pi(h)-P$ = 0.

$$ Lagrangian = objective function + constraint $$ $$\mathcal{L}(h,P,\lambda) = \phi(h)+P + \lambda(\Pi(h,P))$$ $$\mathcal{L}(h,P,\lambda) = -2h^2+P + \lambda(\pi(h) - P)$$ $$\mathcal{L}(h,P,\lambda) = -2h^2+P + \lambda(120-2(h-10)^2 - P)$$ $$\frac{\partial\mathcal{L}}{\partial h} = -4h -4\lambda(h-10) = 0 \rightarrow \lambda = -\frac{h}{h-10}$$ $$\frac{\partial\mathcal{L}}{\partial P} = 1 -\lambda = 0 \rightarrow \lambda = 1$$ $$\frac{\partial\mathcal{L}}{\partial \lambda} = 120-2(h-10)^2 - P = 0$$

When we set both $\lambda$ equal to each other we get $1=\frac{h}{h-10} \rightarrow 2h=5$. Plugging this into $\frac{\partial\mathcal{L}}{\partial \lambda}$ we get $120-2((5)-10)^2 - P=0 \rightarrow 70=P$

Thus $h_p^c = 5$ and $P^c= 70$

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    $\begingroup$ Hint for 1: By having the ability to reject an offer, the consumer is guaranteed to have a utility of at least 0. So the firm's offer should try to maximize its profit subject to this constraint. Part 2 can be approached with a similar reasoning. $\endgroup$
    – Herr K.
    Apr 7 at 16:32
  • $\begingroup$ Hi, Welcome to Economics SE! we have a policy regarding homework questions where users must demonstrate effort using text, markdown syntax and/or images space provided in the text of his question submission for the site. Using only images as proof of work is insufficient. for more details see:economics.meta.stackexchange.com/questions/1465/… $\endgroup$
    – 1muflon1
    Apr 7 at 18:04
  • $\begingroup$ @1muflon1 I have edited the question to reflect my attempts at solving the problem so far. I hope it now meets the requisite standard and can be reopened. $\endgroup$
    – asd7
    Apr 7 at 22:05
  • $\begingroup$ yes, I reopened it $\endgroup$
    – 1muflon1
    Apr 7 at 22:09
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Your steps look correct. There is one small typo in the first $\frac{\partial \mathcal L}{\partial h}$: the term $-4h(h-10)$ should have been $-4(h-10)$.

The results also look reasonable: Regardless of who makes the offer, the Pareto optimal level of $h$ is produced. Given how the bargaining procedures are structured, whoever gets to make the offer gets to keep the surplus of $20$.


P.S. There is an issue in part 2 of the question. It says: "If the firm rejects the offer it cannot produce so the firm's profit is 0." This statement is a contradiction. If the firm cannot produce ($h=0$), then profit would be $-80<0$; if its profit is $0$, then it must produce some positive amount of $h$. But this lack of rigor is on whoever wrote the question, not you.

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  • $\begingroup$ Thank you, I edited the typo. $\endgroup$
    – asd7
    Apr 8 at 6:58
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Given that you already know the welfare maximizing level $h=5$ from your previous question, another approach would be to just consider that any optimal take-it-or-leave-it offer can be divided into two steps: First, maximize total surplus by setting $h=5$. Second, extract all surplus by maximizing the price subject to the other's participation constraint. The latter translates to setting utility or profit, respectively, to zero. The optimal price offer then follows immediately.

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