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:

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

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 ${\begin{aligned}R'(x)&=30-{\frac {x}{2}}-70-(100-x)(-1)\\&=60-{\frac {3x}{2}}\end{aligned}}$

Now we want to maximize the revenue with respect to the number of bus tickets we sell; hence we let $R'(x)=0$ 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.