A Florida juice company completes the preparation of its products by sterilizing, filling, and labeling bottles. Each case of orange juice requires 9 minutes for sterilizing, 6 minutes for filling, and 1 minute for labeling. Each case of grapefruit requires 4 minutes for filling, 10 minutes for sterilizing, and 2 minutes for labeling. Each case of tomato juice requires 1 minute for labeling, 4 minutes for filling, and 12 minutes for sterilizing. The company runs the sterilizing machine for 398 minutes, the filling machine for 164 minutes, and the labeling machine for 58 minutes, how many cases of each type of juice are prepared?
1. Carefully identify your variables and write the equations that need to be solved.
2. Solve the system by the Gauss-Jordan elimination method.

Answer:

1.

R= Number of orange juice cases.

G = Number of greapefruit juice cases.

T = Number of tomato juice cases.

Equations:

9R+10G+12T=398

6R+4G+4T=164

R+2G+T=58

2.

R=6

G=20

T=12

Step-by-step explanation:

1.

The problem is talking about three types of products and it wants us to find how many of each the factory prepares. Since this is the data we need to know, then we set them to be our variables:

R= Number of orange juice cases.

G = Number of greapefruit juice cases.

T = Number of tomato juice cases.

Next, we can use the provided information to build our equations. First, we start with the Sterilizing machine equation: 9 minutes for orange juice, 10 minutes for grape juice and 12 minutes for tomato juice. So we use the sterilizing values for each of the product, to build our first equation:

9R+10G+12T=398

next, we build the filling machine equation, so like on the previous equation, we use the provided data for filling to build the second equation:

6R+4G+4T=164

and finally we build the labeling machine equation so we get:

R+2G+T=58

so we need to solve the following system of equations:

9R+10G+12T=398

6R+4G+4T=164

R+2G+T=58

2.

The very first thing we need to do in order to solve this problem by using the Gauss-Jordan elimination method is to build our matrix based on the system of equations we got on the previous part.

[tex]\left[\begin{array}{cccc}9&10&12&398\\6&4&4&164\\1&2&1&58\\\end{array}\right][/tex]

The idea is to end up with an identity matrix on the first three columns, that will directly give us the answer to the system of equations. So we can start by dividing the first row into 9: [tex]\frac{R_{1}}{9}[/tex] So we get:

[tex]\left[\begin{array}{cccc}1&\frac{10}{9}&\frac{4}{3}&\frac{398}{9}\\6&4&4&164\\1&2&1&58\\\end{array}\right][/tex]

Next, we can multiply the first row by -6 and add it to the second row to get the new second row. [tex]-6R_{1}+R_{2}[/tex]

so we get:

-6   -20/3   -8   -796/3

6     4          4      164

---------------------------------

0    -8/3    -4    -304/3

our matrix now looks like this:

[tex]\left[\begin{array}{cccc}1&\frac{10}{9}&\frac{4}{3}&\frac{398}{9}\\0&-\frac{8}{3}&-4&-\frac{304}{3}\\1&2&1&58\\\end{array}\right][/tex]

Next, we can subtract R3 from R1 so we get:

1   10/9   4/3   398/9

-1   -2     -1       -58

------------------------------

0    -8/9   1/3   -124/9

So the matrix looks like this now:

[tex]\left[\begin{array}{cccc}1&\frac{10}{9}&\frac{4}{3}&\frac{398}{9}\\0&-\frac{8}{3}&-4&-\frac{304}{3}\\0&-\frac{8}{9}&\frac{1}{3}&-\frac{124}{9}\\\end{array}\right][/tex]

Now, we can multiply the second row by -3/8 so we get:

[tex]\left[\begin{array}{cccc}1&\frac{10}{9}&\frac{4}{3}&\frac{398}{9}\\0&1&\frac{3}{2}&38\\0&-\frac{8}{9}&\frac{1}{3}&-\frac{124}{9}\\\end{array}\right][/tex]

Now, we can subtract: [tex]R_{1}-\frac{10}{9}R_{2}[/tex] so we get:

0   -10/9   -5/3  -380/9

1      10/9    4/3   398/9

------------------------------

1        0       -1/3   2

So the matrix will now look like this:

[tex]\left[\begin{array}{cccc}1&0&-\frac{1}{3}&2\\0&1&\frac{3}{2}&38\\0&-\frac{8}{9}&\frac{1}{3}&-\frac{124}{9}\\\end{array}\right][/tex]

Next, we do the following operation: [tex]R_{3}+\frac{8}{9}R_{2}[/tex]

0      8/9      4/3 304/9

0      -8/9     1/3   -124/9

------------------------------

0       0         5/3   20

So our matrix will now look like this:

[tex]\left[\begin{array}{cccc}1&0&-\frac{1}{3}&2\\0&1&\frac{3}{2}&38\\0&0&\frac{5}{3}&20\\\end{array}\right][/tex]

We next multiply R3 by 3/5 so we get:

[tex]\left[\begin{array}{cccc}1&0&-\frac{1}{3}&2\\0&1&\frac{3}{2}&38\\0&0&1&12\\\end{array}\right][/tex]

and now we do: [tex]\frac{R_{3}}{3}+R_{1}[/tex]

0         0      1/3       4

1           0     -1/3      2

------------------------------

1       0          0        6

So our matrix will now look like this:

[tex]\left[\begin{array}{cccc}1&0&0&6\\0&1&\frac{3}{2}&38\\0&0&1&12\\\end{array}\right][/tex]

and now we do the following: [tex]R_{2}-\frac{3}{2}R_{3}[/tex]

so we get:

0         0      -3/2     -18

0         1       3/2      38

------------------------------

0        1          0        20

for our final matrix to be:

[tex]\left[\begin{array}{cccc}1&0&0&6\\0&1&0&20\\0&0&1&12\\\end{array}\right][/tex]

so now we can retrieve the corresponding answers:

R=6

G=20

T=12

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ACTIVIDAD 1 : Según el boletín #329, d/f 10/02/2021 del Ministerio de Salud Pública, el número total de casos infectados hasta esa fecha en el Distrito Nacional, era el doble más 12,585 casos más que los que había en la provincia de la Santiago. ¿Cuántos contagiados hay en cada una de las provincias si entre las dos suman 89,766 casos?
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