<blockquote>
<p>Suppose two goods <span class="math-container">$1$</span> and <span class="math-container">$2$</span> are produced by a monopolist firm. The market demands are <span class="math-container">$$q_1(p_1, p_2) = a-p_1+bp_2 \tag{1}$$</span>
<span class="math-container">$$q_2(p_1, p_2) = a-p_2+bp_1 \tag{2}$$</span></p>
<p>and the cost functions are <span class="math-container">$c_i(q_i) = cq_i$</span>.</p>
<p>What's the profit maximizing price of the monopolist?</p>
</blockquote>
<p>The profit function is given by <span class="math-container">$\pi(p_1, p_2, q_1, q_2) = (p_1-c)q_1 + (p_2-c)q_2$</span>. We need to solve the following maximization problem: <span class="math-container">$$\max[\pi(p_1, p_2, q_1, q_2)] \text{ subject to } q_1 = a-p_1+bp_2 \text{ and } q_2 = a-p_2+bp_1$$</span></p>
<p>The Lagrangian is <span class="math-container">$\mathcal{L}(p_1, p_2, q_1, q_2) = \pi + \lambda_1(\cdot) + \lambda_2(\cdot)$</span> which upon solving <span class="math-container">$\left(\frac{\partial \mathcal{L}}{\partial p_i} = \frac{\partial \mathcal{L}}{\partial q_i} = 0\right)$</span> gives us nothing more than plugging in equations (1) and (2). Where am I going wrong?</p>
<hr />
<p>I am not looking for ways to solve this: I am aware that differentiating <span class="math-container">$\pi$</span> wrt <span class="math-container">$p_1$</span> and <span class="math-container">$p_2$</span> will give the answer. I am curious why the constrained maximization set-up isn't working here.</p>

CC BY-SA 4.0
2022-10-06T16:57:31.720Z
what do you mean it does not work? It seems that it is working. Constrained optimization problem is not in principle different from substitution method. It is just a more convenient way to solve optimization problems where direct substitution is too difficult, you should generally get the same solution whether you substitute or use lagrangian, similarly as calculus yields the same top of parabola as solving for vertex of parabola algebraically
1muflon1
CC BY-SA 4.0
2022-10-06T17:36:20.013Z
@1muflon1 You're right. Yes, it finds the local optimum values while ensuring the "conditions" (which is essentially substitution). I computed it incorrectly and so I didn't get anything beyond vague values of $\lambda$ and the two conditions. Thank you for confirming.
Xelina
CC BY-SA 4.0
2022-10-06T18:27:33.397Z
I’m voting to close this question because it was based on a miscalculation.
Giskard
<microeconomics><monopoly>
Profit maximization in a monopoly
Xelina
# Profit maximization in a monopoly

(Asked by: username_0 on 2022-10-06)

> Suppose two goods $1$ and $2$ are produced by a monopolist firm. The market demands are $$q\_1(p\_1, p\_2) = a-p\_1+bp\_2 \tag{1}$$
> $$q\_2(p\_1, p\_2) = a-p\_2+bp\_1 \tag{2}$$
>
> and the cost functions are $c\_i(q\_i) = cq\_i$.
>
> What's the profit maximizing price of the monopolist?

The profit function is given by $\pi(p\_1, p\_2, q\_1, q\_2) = (p\_1-c)q\_1 + (p\_2-c)q\_2$. We need to solve the following maximization problem: $$\max[\pi(p\_1, p\_2, q\_1, q\_2)] \text{ subject to } q\_1 = a-p\_1+bp\_2 \text{ and } q\_2 = a-p\_2+bp\_1$$

The Lagrangian is $\mathcal{L}(p\_1, p\_2, q\_1, q\_2) = \pi + \lambda\_1(\cdot) + \lambda\_2(\cdot)$ which upon solving $\left(\frac{\partial \mathcal{L}}{\partial p\_i} = \frac{\partial \mathcal{L}}{\partial q\_i} = 0\right)$ gives us nothing more than plugging in equations (1) and (2). Where am I going wrong?

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I am not looking for ways to solve this: I am aware that differentiating $\pi$ wrt $p\_1$ and $p\_2$ will give the answer. I am curious why the constrained maximization set-up isn't working here.

**Tags:** microeconomics, monopoly

> **Comments:**
> **username_1**: what do you mean it does not work? It seems that it is working. Constrained optimization problem is not in principle different from substitution method. It is just a more convenient way to solve optimization problems where direct substitution is too difficult, you should generally get the same solution whether you substitute or use lagrangian, similarly as calculus yields the same top of parabola as solving for vertex of parabola algebraically  
> **username_0**: @username_1 You're right. Yes, it finds the local optimum values while ensuring the "conditions" (which is essentially substitution). I computed it incorrectly and so I didn't get anything beyond vague values of $\lambda$ and the two conditions. Thank you for confirming.  
> **username_2**: I’m voting to close this question because it was based on a miscalculation.  

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## 0 answers

