<p>we need to solve:
<span class="math-container">$$\begin{align}
\max \quad &amp; \pi= \pi_1+\frac{\pi_2}{1+r}=p_1q_1+p_2q_2 \\
\textrm{s.t.} \quad &amp; q_1=150-p_1, \quad q_2=160-p_2 \\
&amp; 0\leq q_1+q_2 \leq 100
\end{align}$$</span>
rewriting in terms of prices
<span class="math-container">$$\begin{align}
\max_{p_1,p_2} \quad &amp;\pi=p_1(150-p_1)+\frac{p_2(160-p_2)}{1.1} \\
\textrm{s.t.} \quad &amp; 210\leq p_1+p_2 \leq 310
\end{align}$$</span></p>
<p>you can check that <span class="math-container">$\pi(p_1,p_2)=p_1(150-p_1)+\frac{p_2(160-p_2)}{1.1}$</span> is maximum at <span class="math-container">$(p_1,p_2)=(75,80)$</span> but <span class="math-container">$p_1+p_2=75+80=155\not \geq 210$</span>, then it must be the case that at the optimum we have <span class="math-container">$p_1+p_2=210$</span></p>
<p>so, the problem we need to solve is:
<span class="math-container">$$\begin{align} 
\max_{p_1,p_2\geq 0} \quad &amp; \pi(p_1,p_2)=p_1(150-p_1)+\frac{p_2(160-p_2)}{1.1}\\
\textrm{s.t.} \quad &amp; p_1+p_2=210  \end{align}$$</span></p>
<p>we can use Lagrangian method to solve the above problem, so the optimal price must satisfy the first order conditions:
<span class="math-container">$$\begin{eqnarray}
\frac{\partial \pi / \partial p_1}{\partial \pi / \partial p_2}=1 \tag{1}\\
p_1+p_2=210 \tag{2}
\end{eqnarray}$$</span></p>
<p>from <span class="math-container">$(1): \quad 150-2p_1=\frac{160-2p_2}{1.1} \implies \boxed{p_2=\frac{11}{10}p_1-\frac{5}{2}}$</span></p>

CC BY-SA 4.0
2023-05-21T11:12:04.597Z
you shouldnt provide answers to a homework questions that show no attempt
1muflon1
CC BY-SA 4.0
2023-05-21T11:12:34.833Z
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
mynameparv
CC BY-SA 4.0
2023-05-21T11:13:35.307Z
@1muflon1 oh sorry, I didn't know that, should I take it down?
mynameparv
CC BY-SA 4.0
2023-05-21T15:35:38.420Z
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.
Ernst Chen
CC BY-SA 4.0
2023-05-21T19:01:57.747Z
@mynameparv just please don’t do it next time
1muflon1
CC BY-SA 4.0
2023-05-21T11:10:10.630Z
2023-05-21T11:10:10.630Z
mynameparv
<p><strong>hint:</strong> hopefully you learned by now that if there are two options, lets say <span class="math-container">$x$</span> and <span class="math-container">$y$</span> rational people will optimally  choose <span class="math-container">$x$</span> and <span class="math-container">$y$</span> such that marginal benefit of <span class="math-container">$x$</span> is equal to marginal benefit of <span class="math-container">$y$</span>.</p>
<p>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?</p>

CC BY-SA 4.0
2023-05-21T09:05:12.827Z
2023-05-21T09:05:12.827Z
1muflon1
<p>There are <span class="math-container">$100$</span> 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:</p>
<hr />
<p><span class="math-container">$q_{1}=150−p_{1}$</span>,
<span class="math-container">$q_{2}=160−p_{2}$</span></p>
<p>where <span class="math-container">$p_{t}$</span> and <span class="math-container">$q_{t}$</span> 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.</p>
<p>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 <span class="math-container">$p_{2}$</span> in terms of <span class="math-container">$p_{1}$</span>?</p>

<microeconomics><supply-and-demand><interest-rate><nash-equilibrium><general-equilibrium>
How to find the equilibrium amount of $p_{2}$ in terms of $p_{1}$?
Ernst Chen
# How to find the equilibrium amount of $p_{2}$ in terms of $p_{1}$?

(Asked by: username_0 on 2023-05-21)

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:

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$q\_{1}=150−p\_{1}$,
$q\_{2}=160−p\_{2}$

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 $p\_{1}$?

**Tags:** microeconomics, supply-and-demand, interest-rate, nash-equilibrium, general-equilibrium


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## 2 answers

### Answer by username_1 on 2023-05-21. Score: 0 | Accepted answer

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 $(1): \quad 150-2p\_1=\frac{160-2p\_2}{1.1} \implies \boxed{p\_2=\frac{11}{10}p\_1-\frac{5}{2}}$

> **Comments:**
> **username_2**: you shouldnt provide answers to a homework questions that show no attempt  
> **username_1**: 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  
> **username_1**: @username_2 oh sorry, I didn't know that, should I take it down?  
> **username_0**: 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.  
> **username_2**: @username_1 just please don’t do it next time  

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### Answer by username_2 on 2023-05-21. Score: 0

**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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