Science:Math Exam Resources/Courses/MATH104/December 2014/Question 04/Solution 1

We know that if we sell $x$ bus tour tickets, we sell $y=100-x$ train tour tickets. Let us now make the revenue function:

$\begin{array}{rl}R& =x(30-\frac{x}{4})+y(70-\frac{y}{2})\\ & =x(30-\frac{x}{4})+(100-x)(70-\frac{100-x}{2})\\ & =30x-\frac{x^2}{4}+70(100-x)-\frac{(100-x{)}^2}{2}\\ \end{array}$

The maximum of $R$ can either be at the boundary of the interval or a local maximum. For $x=0$ we find $R(0)=2000$ and for $x=100$ we obtain $R(100)=2500$. To find the critical point we take the derivative of $R$ with respect to $x$. Being careful with the chain rule in the last summand we obtain $\begin{array}{rl}{R}^{\prime }(x)& =30-\frac{x}{2}-70-(100-x)(-1)\\ & =60-\frac{3x}{2}\end{array}$

Now we want to maximize the revenue with respect to the number of bus tickets we sell; hence we let $R^{\prime }(x)=0$ and determine $x$. We obtain that $x=40$ and $R(40)=3200$. This means that if we sell 40 bus tour tickets and 60 train tour tickets, we maximize the revenue.