Science:Math Exam Resources/Courses/MATH104/December 2010/Question 05

MATH104 December 2010

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Question 05

A small manufacturer wholesales leather jackets to a number of specialty stores. The monthly demand from these stores for the jackets is described by the demand equation $\displaystyle {}p=400-50q.$ Here p is the wholesale price, in dollars per jacket, and q is the monthly demand, in thousands of jackets. Note that the demand equation makes no sense if $q\geq {8}$. The manufacturer's marginal cost is given by the equation, ${\frac {{\textrm {d}}C}{{\textrm {d}}q}}={\frac {800}{q+5}}.$ Determine the number of jackets that must be sold per month to maximize monthly profit. You do not need to justify that your answer provides the maximal profit.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
This is an optimization problem. What is the objective function (the quantity that you want to maximize or minimize)? What are the constraint(s)?

Hint 2
Note that you are not given the cost function in this problem, but only its derivative. Do not try to find the cost function itself. You'll only need the derivative in order to solve the problem.

Hint 3
State your objective function, the profit, as a function of the number of jackets sold per week, $q.$ Look for critical points of this function in order to find its absolute maximum.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. The quantity we want to minimize (the "objective function"), is the profit, $P$. The profit is equal to the revenue, $pq,$ minus the cost, $C(q)$. $\displaystyle {}P=pq-C(q).$ Our constraint is the demand equation, which gives us a relationship between $p$ and $q$: $\displaystyle {}p=400-50q.$ We now use the constraint to express the objective function $P$ in terms of one variable only. Since $C$ is already expressed in terms of $q$, it will be easiest to express $P$ as a function of $q$ rather than $p$. $\displaystyle {}P(q)=(400-50q)q-C(q)=400q-50q^{2}-C(q)$ The domain of interest is $[0,8]$, because we cannot produce a negative quantity of jackets, and in order to sell more than 8 jackets per month, we'd actually have to start paying people to take them (according to the demand equation). In order to find the absolute maximum value of $P(q)$ in this interval, we start by looking for its critical points. $P'(q)=400-100q-C'(q)=400-100q-{\frac {800}{q+5}}$ This exists everywhere inside the domain. In order to search for its zeros, it is helpful to take out a factor of ${\frac {100}{q+5}}$ (which is never equal to zero), so that we are left with a polynomial in $q$. (Hopefully we'll be able to find the roots of that polynomial.) ${\begin{aligned}P'(q)&={\frac {100}{q+5}}\left((4-q)(q+5)-8\right)\\&={\frac {100}{q+5}}\left(4q+20-q^{2}-5q-8\right)\\&=-{\frac {100}{q+5}}\left(q^{2}+q-12\right)\\&=-{\frac {100}{q+5}}(q+4)(q-3)\end{aligned}}$ This is equal to zero when $q$ is equal to 3 or -4, but only 3 lies in the domain of interest. Although it is not necessary, it is easy to check that the profit is actually maximized when $q=3$ because when $q<3,$ $P'(q)>0$, and when $q>3,$ $P'(q)<0$. (Hence, the profit function is increasing when $q<3$ and decreasing when $q>3$.) Therefore, the profit is maximized when 3000 jackets are sold per month. Also, notice that taking the derivative of profit and setting it to zero, is equivalent to finding the point at which the marginal profit is zero or when the marginal revenue is equal to the marginal cost

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