# Quasi fixed costs - two technologies

If a firm has the choice of using two production technologies to produce the same output:

A, featuring a quasi-fixed cost of 50 (i.e. 50 for all q > 0, 0 when q = 0), then a variable cost of 5q

B, which has no fixed costs but a variable cost of q^2 (so MC = 2q)

Then would the firm's optimal strategy be to continue using B until the marginal cost of using B equals 50, and only then consider using A?

• On the one hand, I can see why this would work - it's like the firm is considering what technology to use for q = 1, 2, 3, ... and since the cost to get A started is so high, you'd use B until its marginal cost is high enough to make the firm indifferent between producing the extra unit with B, or switching A on and incurring the fixed cost. But if the MC of using A is very low, and it's just the high quasi-fixed cost deterring the firm from using A, it seems like a poor decision to hold off using A for that long. (Apologies if this doesn't make sense...) Jan 11, 2022 at 3:23
• Does the firm want to minimize cost or marginal cost? Jan 11, 2022 at 4:01
• I realise that the former is true. Would it then be correct to say that marginal costs don't really matter for the firm's cost minimisation problem? Jan 12, 2022 at 17:34
• That depends on what you mean by "don't really matter". Clearly marginal costs and (total) costs are related. Jan 12, 2022 at 19:50
• At every level of $q$ means an infinite number of possibilities. How do you check each without using some optimization trick? These usually rely on the first derivative, marginal cost, as was also the case in tdm's answer. Jan 13, 2022 at 2:32

We need to solve the following problem:

$$\min_{q_A, q_B}\left\{50 \times I_{q_A > 0} + 5 q_A + q_B^2\right\} \text{ s.t. } q_A + q_B = q \text{ and } q_A, q_B \ge 0$$ We split up according to whether $$q_A = 0$$ or $$q_A > 0$$.

• If $$q_A = 0$$ then $$q_B = q$$ and we have a total cost of $$q^2$$.

• If $$q_A > 0$$ then we will choose $$q_A$$ and $$q_B$$ such that (setting up the K-T conditions):

\begin{align*} &5 - \lambda = 0\\ &2 q_B - \lambda - \mu = 0\\ &\mu q_B = 0\\ &q_A + q_B = q \end{align*} Where $$\lambda$$ and $$\mu$$ are the Lagrange mulitpliers to the constraints $$q_A + q_B = q$$ and $$q_B \ge 0$$. (the restriction $$q_A > 0$$ is supposed to hold by assumption). We get that:

• If $$q_B = 0$$ then $$q_A = q$$ and we have a total cost of $$50 + 5 q$$.

• If $$q_B > 0$$ then $$\mu = 0$$ and we get the solution $$q_B = \frac{5}{2}$$ and $$q_A = q - \frac{5}{2}$$. Notice that this can only be a solution for $$q > \frac{5}{2}$$ (as $$q_A > 0$$). In this case, total costs are $$50 + 5(q - \frac{5}{2}) + \frac{25}{4} = 5 q + 50 - \frac{25}{4}$$

So we end up with 3 candidate solutions:

• $$q_A = 0, q_B = q$$ and total costs are $$q^2$$.
• $$q_A = q, q_B = 0$$ and total costs are $$5q + 50$$
• If $$q > \frac{5}{2}$$ we have a potential solution with $$q_A = q - \frac{5}{2}, q_B = \frac{5}{2}$$ and total costs are $$5 q + 50 - \frac{25}{4}$$.

To see which one is best for which levels of $$q$$, notice that the first solution has a higher cost than the second one if: \begin{align*} &q^2 - 5q - 50 \ge 0,\\ \leftrightarrow &q \ge 10 \end{align*} Next, notice that the third solution will always have a lower cost than the second option (in case $$q > \frac{5}{2}$$). and the second option beats the first one only if $$q \ge 10$$. So the second option will never be optimal because $$q \ge 10$$ implies $$q > \frac{5}{2}$$.

We are left to decide between the first and third option. The first option will have a higher cost than the third option if: \begin{align*} &q^2 - 5q - (50 - \frac{25}{4}) \ge 0,\\ \leftrightarrow &q \ge \frac{5}{2}(1 + 2 \sqrt{2}). \end{align*} So we have that:

• if $$q \le \frac{5}{2}(1 + 2 \sqrt{2})$$ then $$q_A = 0$$ and $$q_B = q$$.
• if $$q > \frac{5}{2}(1 + 2 \sqrt{2})$$ then $$q_A = \frac{5}{2}$$ and $$q_B = q - \frac{5}{2}$$.