Q. 81 - Precalculus Enhanced with Graphing Utilities

Question

Production To manufacture an automobile requires painting, drying, and polishing. Epsilon Motor Company produces three types of cars: the Delta, the Beta, and the sigma. Each Delta requires 10 hours (hr) for painting, 3 hr for drying, and 2 hr for polishing. A Beta requires 16 hr for painting, 5 hr for drying, and 3 hr for polishing, and a Sigma requires 8 hr for painting, 2 hr for drying, and 1 hr for polishing. If the company has 240 hr for painting, 69 hr for drying, and 41 hr for polishing per month, how many of each type of car are produced?

Step-by-Step Solution

Answer

The number of Delta cars are $8,$ the number of Beta cars are $5,$ and the number of Sigma cars are $10.$

Step 1. Given Information

A company produces three types of cars: the Delta, the Beta, and the sigma. Each Delta requires 10 hours (hr) for painting, 3 hr for drying, and 2 hr for polishing, a Beta requires 16 hr for painting, 5 hr for drying, and 3 hr for polishing, and a Sigma requires 8 hr for painting, 2 hr for drying, and 1 hr for polishing. The company has 240 hr for painting, 69 hr for drying, and 41 hr for polishing per month. We have to find the number of each type of car are produced.

Step 2. Forming a system of equations

Let a be the number of Delta, b be the number of Beta, and c be the number of Sigma.

The system of equations we get are,

$\left\{\begin{array}{c}10a+16b+8c=240\\ 3a+5b+2c=69\\ 2a+3b+c=41\end{array}\right.$

Step 3. Solve

Let's write the augmented matrix that represents the system 

$$\begin{gathered} \left[\begin{array}{ccccc}10& 16& 8& |& 240\\ 3& 5& 2& |& 69\\ 2& 3& 1& |& 41\end{array}\right] \\ {R}_1=r_1-3r_3 \\ \left[\begin{array}{ccccc}1& 1& 2& |& 33\\ 3& 5& 2& |& 69\\ 2& 3& 1& |& 41\end{array}\right] \\ {R}_2=r_2-3r_1 \\ \left[\begin{array}{ccccc}1& 1& 2& |& 33\\ 0& 2& -4& |& -30\\ 2& 3& 1& |& 41\end{array}\right] \\ {R}_3=r_3-2r_1 \\ \left[\begin{array}{ccccc}1& 1& 2& |& 33\\ 0& 2& -4& |& -30\\ 0& 1& -3& |& -25\end{array}\right] \end{gathered}$$

Step 4. Solve

By proceeding with the calculation further

$$\begin{gathered} R_2=\frac{1}{2}{r}_2 \\ \left[\begin{array}{ccccc}1& 1& 2& |& 33\\ 0& 1& -2& |& -15\\ 0& 1& -3& |& -25\end{array}\right] \\ {R}_3=r_3-r_2 \\ \left[\begin{array}{ccccc}1& 1& 2& |& 33\\ 0& 1& -2& |& -15\\ 0& 0& -1& |& -10\end{array}\right] \\ {R}_3=-1·r_3 \\ \left[\begin{array}{ccccc}1& 1& 2& |& 33\\ 0& 1& -2& |& -15\\ 0& 0& 1& |& 10\end{array}\right] \end{gathered}$$

Step 5. Use row echelon form

The matrix is now in row echelon form. The final matrix represents the system 

$\left\{\begin{array}{c}a+b+2c=33 \left(1\right)\\ b-2c=-15 \left(2\right)\\ c=10 \left(3\right)\end{array}\right.$

Step 6. Solve for b

To find b, we will substitute c in the equation in (2) 

$$\begin{gathered} b-2\left(10\right)=-15 \\ b-20=-15 \\ b=5 \end{gathered}$$

Step 7. Solve for a

To find a, we will substitute b and c in the equation in (1) 

$$\begin{gathered} a+5+2\left(10\right)=33 \\ a+5+20=33 \\ a=8 \end{gathered}$$

Therefore, the number of Delta cars are produced is $8,$ the number of Beta cars are produced is $5,$ and the number of Sigma cars are produced is $10.$