simple arithmatic question Train services on a railway branch line cost $ \$ 1600$ per month to operate. Passengers consists of two cohorts (business passengers and retired holiday makers):
business passengers with aggregated demand $ Q_d=2000-10P$, where $ Q_d$ denotes the number of journeys made per month and $P$ is the price in cents charged for each journey,
while the retired holiday makers has demand function $Q_d=4000-50P$.
How much the railway authority charge if same price is charged for every one?
Answer:
If every passenger is charged equally ($P$ cents), then
$ (2000-10P) \times \text{P cents}+(4000-50P) \times \text{P cents}=\$1600 \\ \Rightarrow 60P^2-6000P+160000=0 \ \  \ $    ($ \because P \text{cent}=\frac{P}{100} \text{dollar}$)
which give no solution of $P$.
Help me
 A: Your attempt at an answer actually shows that it is not possible to get a positive profit when serving both passengers. To finish the work, you need to check it is not worthwhile to serve just the business passengers. Below I show how you would find the profit-maximizing price (which is what you would need to do if a positive profit were possible).

The market demand function $D$ is given by
$$D(P)=\begin{cases}6000-60P&P\leq 80\\ 2000-10P & 80<P\leq 200\end{cases}$$
The price should be chosen to maximize profit. Here there are no variable costs, so if it is worthwhile to run the rail service, then maximizing profit amounts to maximizing revenue.
There are two possibilities: either revenue is maximized at a price $P\leq 80$ or at a price $P>80$.

When $P\leq 80$, revenue is
$$P(6000-60P)$$
This is a quadratic with intercepts at $P=0$ and $P=100$, so the maximizer is $P=50\leq 80$ where revenue is $\$1500$

When $P>80$, revenue is
$$P(2000-10P)$$
This is a quadratic with intercepts at $P=0$ and $P=200$, so the maximizer is at $P=100>80$, where revenue is $\$1000$.

Since neither revenue is more than the fixed cost of operating, the train service should not operate.
A: Hint:  The profit is equal to the price times the number of passengers minus the operating costs
$$
\begin{align}
\mathrm{Profit}&= P\cdot (Q_b(P) + Q_h(P)) - \mathrm{Cost}\\
&=P((2000-10\;P)+(4000-50P))-160000\\
&=-60 P^2+6000 P-160000
\end{align}
$$
where $Q_b(P)$ is the monthly number of business passengers that will use the train if the price is $P$ cents per ticket and $Q_h(P)$ is the monthly number of holiday passengers.
Note that we do not want to find the value of P where the Profit is zero.  We want to maximize the profit.
There are two standard ways to find the best price:  take the derivative and set it to zero or use the fact that the vertex of the parabola $y=ax^2+bx+c$ is at the point $(-b/(2a), c-b^2/(4a))$.

PS: Once you have found the best price $P$, make sure that $Q_h(P)$ and $Q_b(P)$ are both positive.  If they are not, then you will need to make some modifications. Also, you need to check whether the optimal $P$ yields a positive profit.
PPS:  Thanks to Sauhard Sharma for pointing out the error in the first version of the answer.
