Q52. - Algebra 2

Question

A tour bus in the historic district of Savannah, Georgia, serves 300 customers a day. The charge is \(8 per person. The owner estimates that the company would lose 20 passengers a day for each \)1 fare increase. If the company raised their fare to this price, how much daily income should they expect to bring in?

Step-by-Step Solution

Answer

The maximum daily income the company should expect to bring in is $2645. 

Step 1. Given Information.

Given,

Charge is$8 per person

The tour bus serves 300 customers per day

Number of passengers decrease by 20 each day for each $1 increase in fare.

Step 2. Use the concept.

Consider the function $f\left(x\right)=a{x}^2+bx+c,a\ne 0$, the x-coordinate of vertex is $\frac{-b}{2a}$.

 

The graph of $f\left(x\right)=a{x}^2+bx+c,a\ne 0$

• opens up and has a minimum value when $a>0$, and
• opens down and has a maximum value when $a<0$

Step 3. Solution.

Let x = the number of $1 increase in fare

Then  $8+1x$ = Charge per person 

$300-20x$= The number of passengers

Let I(x) = income as a function of x.

 $\begin{array}{l}\stackrel{\text{The Income}}{\overbrace{I\left(x\right)}}=\stackrel{Thenumberofpassengers}{\overbrace{\left(300-20x\right)}}\cdot \stackrel{\text{Charge pe}r\text{ person}}{\overbrace{\left(8+1x\right)}}\\ \Rightarrow I\left(x\right)=\text{2400}+300x-\text{160}x-\text{20}{x}^2\\ \Rightarrow I\left(x\right)=-\text{20}{x}^2+\text{140}x+\text{2400}\end{array}$

The charge that would give the most income for the company is $11.5 and $x=3.5$

Therefore, the maximum daily income the company should expect to bring in is:

  $\begin{array}{l}\stackrel{\text{The Income}}{\overbrace{I\left(x\right)}}=\stackrel{Thenumberofpassengers}{\overbrace{\left(300-20x\right)}}\cdot \stackrel{\text{Charge pe}r\text{ person}}{\overbrace{\left(8+1x\right)}}\\ \Rightarrow I\left(3.5\right)=\left(300-20\cdot 3.5\right)\cdot \left(11.5\right)\\ \Rightarrow I\left(3.5\right)=230\cdot \left(11.5\right)\\ \Rightarrow I\left(3.5\right)=2645\end{array}$