Optimal Price and Quantity of Externality

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.
   $$ Lagrangian = objective function + constraint $$
   $$mathcal{L}(h,P,lambda) = Pi(h,P) + lambda(phi(h)+P)$$
   $$mathcal{L}(h,P,lambda) = pi(h) – P + lambda(-2h^2+P)$$
   $$mathcal{L}(h,P,lambda) = 120-2(h-10)^2 – P + lambda(-2h^2+P)$$
   $$frac{partialmathcal{L}}{partial h} = -4(h-10)-4hlambda = 0 rightarrow lambda = frac{-4h+40}{4h}$$
   $$frac{partialmathcal{L}}{partial P} = -1 + lambda = 0 rightarrow lambda = 1$$
   $$frac{partialmathcal{L}}{partial lambda} = -2h^2+P = 0$$

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

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

2. 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.

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

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

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