Science:Math Exam Resources/Courses/MATH104/December 2010/Question 05/Solution 1

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