Chapter 7, Problem 74RE

Solution

To calculate: The number of tickets of each type were sold if the total revenue form the sale of tickets was $\$429$ .

The point is $\left(\frac{3}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$ .

Given Information:

At Whetstone High School 452 tickets were sold for the first basketball game. There were two ticket prices: $\$0.75$ for students and $\$2.00$ for non-students.

Calculation:

Consider the given point,

Suppose that x  be the number of student tickets and  y  be the number of nonstudent tickets.

Set up the equations. There were 452 tickets sold so:

  $x+y=452$

The ticket sales is $429 so:

  $0.75x+2.00y=429$

Write the system of the equation the matrix of the equation as $AX=B$ .

  $A=\left[\begin{array}{cc}1& 1\\ 0.75& 2.00\end{array}\right],X=\left[\begin{array}{c}x\\ y\end{array}\right],\text{ and }B=\left[\begin{array}{c}452\\ 429\end{array}\right]$

Find the inverse of the matrix $A$ .

  $\begin{array}{c}{A}^{-1}={\left[\begin{array}{cc}1& 1\\ 0.75& 2.00\end{array}\right]}^{-1}\\ =\frac{1}{2-0.75}\left[\begin{array}{cc}2.00& -1\\ -0.75& 1\end{array}\right]\\ =\frac{1}{0.25}\left[\begin{array}{cc}2.00& -1\\ -0.75& 1\end{array}\right]\end{array}$

And,

  $\begin{array}{c}X=A^{-1}B\\ =\frac{1}{0.25}\left[\begin{array}{cc}2.00& -1\\ -0.75& 1\end{array}\right]\cdot \left[\begin{array}{c}452\\ 429\end{array}\right]\\ =\left[\begin{array}{c}380\\ 72\end{array}\right]\end{array}$

Hence, the number of tickets sold are 380 student tickets and 72 nonstudent tickets.