Q51. - 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. What charge would give the most income for the company?

Step-by-Step Solution

Answer

The charge that would give the most income for the company is $11.5.

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.

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

In the function$I\left(x\right)=-\text{20}{x}^2+\text{140}x+\text{2400}$, we have

Here,  $\begin{array}{l}a=-20<0\\ b=140\end{array}$

So, the graph opens down and has a maximum value.

The maximum value of the function is the y-coordinate of the vertex.

The x-coordinate of the vertex is

  $\begin{array}{l}\frac{-140}{2(-20)}\\ =\mathrm{3.5..........}\left[a=-20,b=140\right]\end{array}$

So, the tour bus must make 3.5 times of $1  increase in fare.

Therefore, the charge that would give the most income for the company is:

 $\begin{array}{l}8+3.5\\ =11.5\end{array}$