Q. 86 - Precalculus Enhanced with Graphing Utilities

Question

Computing Profit; Rizza Ford has two locations, one in the city and the other in the suburbs. In January, the city location sold $400$ subcompacts, $250$ intermediate-size cars, and $50$ SUVs; in February, it sold $350$ subcompacts, $100$ intermediates, and $30$ SUVs. At the suburban location in January, $450$ subcompacts, $200$ intermediates, and $140$ SUVs were sold. In February, the suburban location sold $350$ subcompacts, $300$ intermediates, and $100$ SUVs.

(a) Find $2$ by $3$ matrices that summarize the sales data for each location for January and February (one matrix for each month).

(b) Use matrix addition to obtain total sales for the $2$-month period.

(c) The profit on each kind of car is $\backslash (100$ per subcompact, $\backslash )150$ per intermediate, and $\$200$ per SUV. Find a $3$ by $1$ matrix representing this profit.

(d) Multiply the matrices found in parts (b) and (c) to get a $2$ by $1$ matrix showing the profit at each location 

Step-by-Step Solution

Answer

(a) The $2$ by $3$ matrix that summarize the sales data for each location for;

 January $=\left[\begin{array}{ccc}400& 250& 50\\ 350& 100& 30\end{array}\right]$

 February $=\left[\begin{array}{ccc}450& 200& 140\\ 350& 300& 100\end{array}\right]$ 

(b) The total sales for the two-month period$=\left[\begin{array}{ccc}850& 450& 190\\ 700& 400& 130\end{array}\right]$

(c) The $3$ by $1$ matrix representing the profit on each kind of car $=\left[\begin{array}{c}100\\ 150\\ 200\end{array}\right]$

(d) The $2$ by $1$ matrix showing the profit at each location $=\left[\begin{array}{c}190500\\ 156000\end{array}\right]$

Step 1. Given data

(a) A company Rizza Ford has two locations, one in the city and the other in the suburbs. 

 In January, the city location sold $400$ subcompacts, $250$ intermediate-size cars, and $50$ SUVs; in February, it sold $350$   subcompacts, $100$ intermediates, and $30$ SUVs. At the suburban location in January, $450$ subcompacts, $200$ intermediates, and $140$ SUVs were sold. In February, the suburban location sold $350$ subcompacts, $300$ intermediates, and $100$ SUVs. 

The matrix summarizes the sales data for each location for each month;

Let the matrix for January be x

$x=\left[\begin{array}{ccc}400& 250& 50\\ 350& 100& 30\end{array}\right]$

Let the matrix for February be y

$y=\left[\begin{array}{ccc}450& 200& 140\\ 350& 300& 100\end{array}\right]$

Step 2. Total sales for the two-month period

(b) To find the total sales for the two-month period, let us add the corresponding entries of the two matrices,

That is,

 $$\begin{gathered} x+y=\left[\begin{array}{ccc}400& 250& 50\\ 350& 100& 30\end{array}\right]+\left[\begin{array}{ccc}450& 200& 140\\ 350& 300& 100\end{array}\right] \\ =\left[\begin{array}{ccc}850& 450& 190\\ 700& 400& 130\end{array}\right] \end{gathered}$$

Step 3. Matrix representing the profit on each kind of car

(c) The matrix that represents the profit on each kind of car, and let it be $z$.

  $z=\left[\begin{array}{c}100\\ 150\\ 200\end{array}\right]$

Step 4.

(d) To find overall profit in each month, we have to multiply $(x+y)$ with$z$.

$$\begin{gathered} (x+y)\times z=\left[\begin{array}{ccc}850& 450& 190\\ 700& 400& 130\end{array}\right]\times \left[\begin{array}{c}100\\ 150\\ 200\end{array}\right] \\ =\left[\begin{array}{c}850\left(100\right)+450\left(150\right)[phone]\mathrm{}\mathrm{}\\ \mathrm{}\left(100\right)+400\left(150\right)+130\left(200\right)\end{array}\right] \\ =\left[\begin{array}{c}190500\\ 156000\end{array}\right] \end{gathered}$$

Therefore, the profit in January is $\$190500$and the profit in February is $\$156000$