Q. 82 - Precalculus Enhanced with Graphing Utilities

Question

Production A Florida juice company completes the preparation of its products by sterilizing, filling, and labeling bottles. Each case of orange juice requires 9 minutes (min) for sterilizing, 6 min for filling, and 1 min for labeling. Each case of grapefruit juice requires 10 min for sterilizing, 4 min for filling, and 2 min for labeling. Each case of tomato juice requires 12 min for sterilizing, 4 min for filling, and 1 min for labeling. If the company runs the sterilizing machine for 398 min, the filling machine for 164 min, and the labeling machine for 58 min, how many cases of each type of juice are prepared?

Step-by-Step Solution

Answer

The number of orange juice prepared is $6,$ the number of grapefruit  juice prepared is $20,$ and the number of tomato juice prepared is $12.$

Step 1. Given Information

There is a company of preparation of juice and each case of orange juice requires 9 minutes (min) for sterilizing, 6 min for filling, and 1 min for labeling, each case of grapefruit juice requires 10 min for sterilizing 4 min for filling, and 2 min for labeling and each case of tomato juice requires 12 min for sterilizing 4 min for filling, and 1 min for labeling. We have to find the number of each type of juice are prepared if the company runs the sterilizing machine for 398 min, the filling machine for 164 min, and the labeling machine for 58 min.

Step 2. Forming a system of equations

Let a be the number of orange juice, b be the number of grapefruit juice, and c be the number of tomato juice.

The system of equations we get are,

$\left\{\begin{array}{c}9a+10b+12c=398\\ 6a+4b+4c=164\\ a+2b+c=58\end{array}\right.$

Step 3. Solve

Let's write the augmented matrix that represents the system  

$$\begin{gathered} \left[\begin{array}{ccccc}9& 10& 12& |& 398\\ 6& 4& 4& |& 164\\ 1& 2& 1& |& 58\end{array}\right] \\ {R}_1=\frac{1}{9}{r}_1 \\ \left[\begin{array}{ccccc}1& \raisebox{1ex}{$10$}\!\left/ \!\raisebox{-1ex}{$9$}\right.& \raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.& |& \raisebox{1ex}{$398$}\!\left/ \!\raisebox{-1ex}{$9$}\right.\\ 6& 4& 4& |& 164\\ 1& 2& 1& |& 58\end{array}\right] \\ {R}_2=r_2-6r_1 \\ \left[\begin{array}{ccccc}1& \raisebox{1ex}{$10$}\!\left/ \!\raisebox{-1ex}{$9$}\right.& \raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.& |& \raisebox{1ex}{$398$}\!\left/ \!\raisebox{-1ex}{$9$}\right.\\ 0& -\raisebox{1ex}{$8$}\!\left/ \!\raisebox{-1ex}{$3$}\right.& -4& |& -\raisebox{1ex}{$304$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\\ 1& 2& 1& |& 58\end{array}\right] \\ {R}_3=r_3-r_1 \\ \left[\begin{array}{ccccc}1& \raisebox{1ex}{$10$}\!\left/ \!\raisebox{-1ex}{$9$}\right.& \raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.& |& \raisebox{1ex}{$398$}\!\left/ \!\raisebox{-1ex}{$9$}\right.\\ 0& -\raisebox{1ex}{$8$}\!\left/ \!\raisebox{-1ex}{$3$}\right.& -4& |& -\raisebox{1ex}{$304$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\\ 0& \raisebox{1ex}{$8$}\!\left/ \!\raisebox{-1ex}{$9$}\right.& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.& |& \raisebox{1ex}{$124$}\!\left/ \!\raisebox{-1ex}{$9$}\right.\end{array}\right] \end{gathered}$$

Step 4. Solve

By proceeding with the calculation further

$$\begin{gathered} R_2=-\frac{3}{8}{R}_2 \\ \left[\begin{array}{ccccc}1& \raisebox{1ex}{$10$}\!\left/ \!\raisebox{-1ex}{$9$}\right.& \raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.& |& \raisebox{1ex}{$398$}\!\left/ \!\raisebox{-1ex}{$9$}\right.\\ 0& 1& \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.& |& 38\\ 0& \raisebox{1ex}{$8$}\!\left/ \!\raisebox{-1ex}{$9$}\right.& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.& |& \raisebox{1ex}{$124$}\!\left/ \!\raisebox{-1ex}{$9$}\right.\end{array}\right] \\ {R}_3=r_3-\frac{8}{9}{r}_2 \\ \left[\begin{array}{ccccc}1& \raisebox{1ex}{$10$}\!\left/ \!\raisebox{-1ex}{$9$}\right.& \raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.& |& \raisebox{1ex}{$398$}\!\left/ \!\raisebox{-1ex}{$9$}\right.\\ 0& 1& \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.& |& 38\\ 0& 0& -\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.& |& -20\end{array}\right] \\ {R}_3=-\frac{3}{5}{r}_3 \\ \left[\begin{array}{ccccc}1& \raisebox{1ex}{$10$}\!\left/ \!\raisebox{-1ex}{$9$}\right.& \raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.& |& \raisebox{1ex}{$398$}\!\left/ \!\raisebox{-1ex}{$9$}\right.\\ 0& 1& \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.& |& 38\\ 0& 0& 1& |& 12\end{array}\right] \end{gathered}$$

Step 5. Use row echelon form

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

$\left\{\begin{array}{c}a+\frac{10}{9}b+\frac{4}{3}c=\frac{398}{9} \left(1\right)\\ b+\frac{3}{2}c=38 \left(2\right)\\ c=12 \left(3\right)\end{array}\right.$

Step 6. Solve for b

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

$$\begin{gathered} b+\frac{3}{2}\left(12\right)=38 \\ b=38-18 \\ b=20 \end{gathered}$$

Step 7. Solve for a

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

$$\begin{gathered} a+\frac{10}{9}\left(20\right)+\frac{4}{3}\left(12\right)=\frac{398}{9} \\ a+\frac{200}{9}+\frac{48}{3}=\frac{398}{9} \\ a=6 \end{gathered}$$

Therefore, the number of orange juice prepared is $6,$ the number of grapefruit juice prepared is $20,$ and the number of tomato juice prepared is $12.$