Constrained optimization problem

Question

So for optimization problems we have only been given scenarios in which we can just solve by doing MRTS= -w/r and the quantity that wants to be produced is stated in the question.

However I was given this question

Here as far as I can understand we can do this via Lagrangian methods, but we have not been taught this yet. As far as I know, the question asking what is the best combination of output from both factories to produce a total of 4 units at the lowest total cost.

In order to do this question you had to set MC₁=MC₂ which you can get via lagrangian methods. However I was told that this condition is also just basic intuition, but I have no idea why having this condition would lead to the smallest total cost. I would gratefully if someone explain the intuition step by step.

Answer 1

Given the cost function of the two plants $c_1(q_1)=10q_1-4q_1^2+q_1^3$ and $c_2(q_2)=10q_2-2q_2^2+q_2^3$, we want to solve the following cost minimisation problem: \begin{eqnarray*}\min_{(q_1,q_2)\in\mathbb{R}^2_+} & c_1(q_1)+c_2(q_2) \\ \text{s.t.} & q_1+q_2=4\end{eqnarray*} Here is the plot of the marginal costs of the two plants, we can observe and compare different possibilities for the solution in the graph below.

If we observe the marginal costs of the two plants in the graph above, we observe that the optimal solution is where two firms operate at the same marginal cost i.e. where $10-8q_1+3q_1^2=10-4q_2+3q_2^2$ and $q_1+q_2=4$ holds. This yields the solution as $q_1^*=\frac{8}{3}$, and $q_2^*=\frac{4}{3}$. In the picture above, the optimal cost is represented by the colored area which is equal to $c_1(\frac{8}{3})+c_2(\frac{4}{3})$.

Answer 2

The intuition is that a firm should produce at the lowest cost possible. Suppose $MC_1<MC_2$, then the firm should let plant 1 produce as long as this inequality holds. Since MC is increasing in both plants, the inequality will eventually be violated. This is when plant 2 becomes the less costly plant and thus should take over production responsibility from plant 1. Eventually, the optimal production arrangement occurs when the two plants have equal MCs.

Formally, the intuition can be demonstrated in the following cost minimization exercise. Let $C(q)$ be the firm-level cost function and $C_i(q_i)$, where $i=1,2$, be the plant-level cost functions. \begin{equation} C(q) = \min_{q_1,q_2}\;C_1(q_1)+C_2(q_2), \qquad\text{such that }q=q_1+q_2. \end{equation} The Lagrangian for the minimization is \begin{equation} \mathcal L(q_1,q_2,\lambda) = C_1(q_2)+C_2(q_2)+\lambda[q-q_1-q_2]. \end{equation} Assuming interior solutions, the FOCs imply \begin{align} \frac{\partial C_i}{\partial q_i} = MC_i(q_i^*) = \lambda, \qquad i=1,2, \end{align} which means $MC_1(q_1^*)=MC_2(q_2^*)$ at the optimum.

Answer 3

A way to show formally (without the Lagrangian method) that for minimum cost we must have $MC_1=MC_2$:

Let $C(q_1)=C_1(q_1)+C_2(q_1)$ where $C_2(q_1)$ is the cost function obtained by substituting $q_2=4-q_1$ in $C_2(q_2)$. A first-order condition for $C$ to be a minimum is:

$$\dfrac{dC}{dq_1}=\dfrac{dC_1}{dq_1}+\dfrac{dC_2}{dq_1}=0\quad\quad(1)$$

But since $q_1=4-q_2$ we have:

$$\dfrac{dq_1}{dq_2}=-1$$

and so:

$$\dfrac{dC_2}{dq_1}=-\dfrac{dC_2}{dq_2}$$

Hence, substituting in (1):

$$\dfrac{dC_1}{dq_1}-\dfrac{dC_2}{dq_2}=MC_1-MC_2=0$$

which implies $MC_1=MC_2$.