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There are $100$ tons of crops remaining to supply for the two months. The crop holders consider whether to sell crops now or one month later. Holders face the demand curve of each period as below:


$q_{1}=150−p_{1}$, $q_{2}=160−p_{2}$

where $p_{t}$ and $q_{t}$ denote the price and quantity of the crop where 1 and 2 stand for this month and the next month, respectively. The monthly interest rate is 10%. Also, Here is assumed that compounding is counted once a month.

The condition of the monthly interest rate baffled me, can someone gives me some ideas to start with this question: find the equilibrium amount of $p_{2}$ in terms of $p_{1}$?

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2 Answers 2

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we need to solve: $$\begin{align} \max \quad & \pi= \pi_1+\frac{\pi_2}{1+r}=p_1q_1+p_2q_2 \\ \textrm{s.t.} \quad & q_1=150-p_1, \quad q_2=160-p_2 \\ & 0\leq q_1+q_2 \leq 100 \end{align}$$ rewriting in terms of prices $$\begin{align} \max_{p_1,p_2} \quad &\pi=p_1(150-p_1)+\frac{p_2(160-p_2)}{1.1} \\ \textrm{s.t.} \quad & 210\leq p_1+p_2 \leq 310 \end{align}$$

you can check that $\pi(p_1,p_2)=p_1(150-p_1)+\frac{p_2(160-p_2)}{1.1}$ is maximum at $(p_1,p_2)=(75,80)$ but $p_1+p_2=75+80=155\not \geq 210$, then it must be the case that at the optimum we have $p_1+p_2=210$

so, the problem we need to solve is: $$\begin{align} \max_{p_1,p_2\geq 0} \quad & \pi(p_1,p_2)=p_1(150-p_1)+\frac{p_2(160-p_2)}{1.1}\\ \textrm{s.t.} \quad & p_1+p_2=210 \end{align}$$

we can use Lagrangian method to solve the above problem, so the optimal price must satisfy the first order conditions: $$\begin{eqnarray} \frac{\partial \pi / \partial p_1}{\partial \pi / \partial p_2}=1 \tag{1}\\ p_1+p_2=210 \tag{2} \end{eqnarray}$$

from $(1): \quad 150-2p_1=\frac{160-2p_2}{1.1} \implies \boxed{p_2=\frac{11}{10}p_1-\frac{5}{2}}$

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  • $\begingroup$ you shouldnt provide answers to a homework questions that show no attempt $\endgroup$
    – 1muflon1
    May 21, 2023 at 11:12
  • $\begingroup$ I think this is what we need to do, but it's possible that I might be incorrect, just decided to share my take regardless $\endgroup$
    – mynameparv
    May 21, 2023 at 11:12
  • $\begingroup$ @1muflon1 oh sorry, I didn't know that, should I take it down? $\endgroup$
    – mynameparv
    May 21, 2023 at 11:13
  • $\begingroup$ It really clarifies my problem. What I did was set $\Pi(p_{1},p_{2})=1.1p_{1}(150-p_{1})+p_{2}(160-p_{2})$ and I stuck when $p_{1}+p{2}$ fails its condition on the constraint. So I thought maybe I misunderstood the "monthly rate" incorrectly. $\endgroup$
    – Ernst Chen
    May 21, 2023 at 15:35
  • $\begingroup$ @mynameparv just please don’t do it next time $\endgroup$
    – 1muflon1
    May 21, 2023 at 19:01
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hint: hopefully you learned by now that if there are two options, lets say $x$ and $y$ rational people will optimally choose $x$ and $y$ such that marginal benefit of $x$ is equal to marginal benefit of $y$.

Now think what is the farmer's benefit of selling today? What is farmer's benefit of selling tomorrow? How does interest rate connect present value to the future?

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