The problem: I am currently working through a course on edX, Principles of Economics with Calculus. The course is structured such that the only prerequisite needed is univariate calculus; accordingly, I have completed the equivalent of Calc I and II, through Taylor series. I've come across the following economic optimization problem, and I've had quite a bit of difficulty reasoning through it. It is worded as follows:

"Imagine that you own a small business and that you have to decide how much of your product to sell every year, and which technology to use to produce it.

Your market research suggests that the price at which you will be able to sell each unit of your product is given by $$200-q,$$ where q denotes the total amount sold over the course of the year.

You have access to two different technologies, which you can use in any feasible combination. The technologies have different cost properties. Producing q units with the first technology costs $$100q,$$ while producing q units with the second technology costs $$q^2.$$

How many units should be produced using the first technology at the profit-maximizing decision?"

My attempt at a solution and concerns: Initially, I made quite a gaffe and did not pay careful attention to the wording of the problem; at the outset, I interpreted it to be asking what the optimal production level would be if we were only using the first technology. I simply wrote the profit function as $$P(q) = q(200-q) - 100q,$$ expanded, differentiated, located the critical points, and verified using the Second Derivative Test. With that process, I arrived at q = 50. After having submitted that answer - which was marked as incorrect - I realized my first error: this question was asking how many particular units should be produced using the first technology if both are used in some configuration.

Knowing the real question being asked, I re-attempted the solution, but my concern is this: how do I - if this is at all possible in the first place - write a revenue function that accounts for the fact that a combination of these two technologies will be used? If q is our total amount being sold, then there will be a particular amount produced using technology 1 - let's call this x - and another amount produced using technology 2, which would be $q-x$. Assigning these variables, I went through the analytic process of optimization once again, but the roadblock I continue to hit is not knowing how to actually differentiate with this involvement of both q and x; should I just assume that q is fixed, and treat it as a constant when differentiating?

Any assistance would be greatly appreciated. Thank you in advance.


2 Answers 2


Since there are two quantities you can set independently this is a two-variable calculus problem. You should call the quantity produced by the first process $q_1$ and the second quantity $q_2$ and then write an expression $P(q_1,q_2)$ for the profit. To find the local extrema of the profit, you take the partial derivative with respect to each parameter and set both of them equal to zero. This will give you two equations in two variables to solve.

It seems a bit odd that it would be a multivariate calculus problem given the prerequisites and background, but that's my best interpretation of the problem.

  • 1
    $\begingroup$ I've seen business calculus courses cover how to compute partial derivatives for this very reason. This edX course is the equivalent of a standard Intermediate Microeconomics course, which is the first hurdle course for economics majors (which are either business students or engineers/scientists/mathematicians who are double majoring). It's hard to teach the course without multiple goods (because then we don't have trade in an economy), so partial derivatives are necessary. That's the most multivariable calculus used, though. $\endgroup$
    – ml0105
    Jan 5, 2017 at 5:48

This is likely too late to help the OP, but it might help others. It is possible to solve this problem without knowledge of partial differentiation.The marginal cost of the first technology is $100$ while the marginal cost of the second technology is $2q$. So long as you want to produce no more than $50$ units, it is better to use only the second technology. Thus if you simply solve the problem: $$\max_q q(200-q)-2q\quad \text{subject to $q\geq 0$}$$ and find that the maximizer is $50$ or less, then you know that it is optimal not to use technology 1 at all.

Even if the problem were slightly different, so that the shortcut above could not be applied, one could still solve the problem without knowing anything about multivariable calculus (or even differentiating at all).

To do so, notice that maximizing profit requires minimizing the cost of producing the profit-maximizing output. This means we can split the problem up into two steps:

  1. Choose the mix of technologies to minimize the cost of producing $q$ (and so obtain minimum cost as a function of $q$)
  2. Substitute the minimum cost function into the profit function and then choose $q$ to maximize profit


The cost when producing $q=q_1+q_2$ where $q_1$ is produced with technology $1$ and $q_2$ with technology $2$ is $$100q_1+q_2^2.$$ We want to solve the problem $$\min_{q_1,q_2} 100q_1+q_2^2\quad \text{subject to $q_1+q_2=q$ and $q_1,q_2\geq 0$}$$ We can substitute for $q_1$ to obtain a one-variable optimization problem:$$\min_{q_2} q_2(q_2-100)+100q\quad \text{ subject to $q_2\in[0,q]$}$$ The maximum of the quadratic in $q_2$ is at $q_2=50$. Thus the cost-minimizing mix is $$(q_1,q_2)=\begin{cases}(0,q) & q\leq 50 \\ (q-50,50)& q\geq 50\end{cases}.$$

Therefore the minimum cost function $C$ is given by $$C(q)=\begin{cases}q^2 & q\leq 50\\ 100(q-50)+2500 & q\geq 50\end{cases}$$ 2.

Now we have to maximize profit. Profit, $\pi$ is given by $$\pi(q)=q(200-q)-C(q)=\begin{cases}2q(100-q) & q\leq 50\\q(100-q)+2500 & q \geq 50.\end{cases}$$ The maximum of the quadratic in each part of the profit function is at $q=50$. Thus the profit-maximizing mix is $(q_1,q_2)=(0,50)$ (and the maximum profit is $5000$).


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .