Answer
The ticket price that will maximize the airline’s monthly revenue is $\$65$ and the maximum monthly revenue is $\$422,500$.
Work Step by Step
The number of passengers $N$ depends on the ticket price, x. In particular the number of passengers is the original number $8000$, minus the number of passengers lost to fare increase.
The number of passengers per month is the original number of passengers, $8000$, minus the decrease due to fare increase.
$N\left( x \right)=8000-100x$
The maximum ticket price is $50+x$.
The monthly revenue $R$ for the route is the number of passengers times the ticket price $50+x$.
$\begin{align}
& R\left( x \right)=\left( 8000-100x \right)\left( 50+x \right) \\
& =400,000-5000x+8000x-100{{x}^{2}} \\
& =400,000+3000x-100{{x}^{2}}
\end{align}$
Thus, the monthly revenue is $R\left( x \right)=-100{{x}^{2}}+3000x+400,000$.
Compare it with the standard equation of quadratic function $f\left( x \right)=a{{x}^{2}}+bx+c$ .
The value of a is $-100$ , b is $3000$ and c is $400,000$ .
Recall that if $a<0$ , then f has a maximum that occurs at $x=\frac{-b}{2a}$ and maximum value is $f\left( \frac{-b}{2a} \right)$
Since, $a=-100$ , $a<0$ , $R$ has a maximum.
Recall that the maximum value is at $x=\frac{-b}{2a}$.
Substitute $-100$ for a and $3000$ for b.
$\begin{align}
& x=\frac{-b}{2a} \\
& =\frac{-3000}{-200} \\
& =15
\end{align}$
And the maximum value is $f\left( \frac{-b}{2a} \right)$
Substitute $15$ for x in $R\left( x \right)=-100{{x}^{2}}+3000x+400,000$.
$\begin{align}
& R\left( x \right)=-100{{x}^{2}}+3000x+400,000 \\
& =-100\left( 225 \right)+3000\left( 15 \right)+400,000 \\
& =422,500
\end{align}$
Hence, the maximum ticket price is $\$65$ and maximum monthly revenue is $\$422,500$.
