The number [tex]\(N\)[/tex] of items produced by a company partly varies as the volume [tex]\(v\)[/tex] of water and partly varies jointly as the number of machines [tex]\(A\)[/tex] and the quantity of raw materials [tex]\(Q\)[/tex].

With 4 machines, 6 silos of raw materials, and 15 cubic meters of water, the company produced 129 boxes of items in a day.

With 19 machines, [tex]\(6\frac{1}{2}\)[/tex] silos of raw materials, and 19 cubic meters of water, the company produced 148 boxes of items.



Answer :

Let's break this problem down step-by-step to find the relationship between the variables [tex]\(N\)[/tex], [tex]\(v\)[/tex], [tex]\(A\)[/tex], and [tex]\(Q\)[/tex] for the production of items.

### Given Data:
1. For the first set of conditions:
- Number of machines ([tex]\(A_1\)[/tex]) = 4
- Number of silos of raw materials ([tex]\(Q_1\)[/tex]) = 6
- Volume of water ([tex]\(v_1\)[/tex]) = 15 m[tex]\(^3\)[/tex]
- Number of items produced ([tex]\(N_1\)[/tex]) = 129

2. For the second set of conditions:
- Number of machines ([tex]\(A_2\)[/tex]) = 19
- Number of silos of raw materials ([tex]\(Q_2\)[/tex]) = 6.5
- Volume of water ([tex]\(v_2\)[/tex]) = 19 m[tex]\(^3\)[/tex]
- Number of items produced ([tex]\(N_2\)[/tex]) = 148

We are given that the number [tex]\(N\)[/tex] of items produced varies partly as the water volume [tex]\(v\)[/tex], and partly as the product of the number of machines [tex]\(A\)[/tex] and the quantity of raw materials [tex]\(Q\)[/tex]. Hence, we can write:

[tex]\[ N = k_1 v + k_2 (A \times Q) \][/tex]

where [tex]\(k_1\)[/tex] and [tex]\(k_2\)[/tex] are constants to be determined.

### Forming the Equations:
Using the given data, we can set up two equations:

For the first set of conditions:
[tex]\[ 129 = k_1 (15) + k_2 (4 \times 6) \][/tex]
[tex]\[ 129 = 15k_1 + 24k_2 \][/tex]

For the second set of conditions:
[tex]\[ 148 = k_1 (19) + k_2 (19 \times 6.5) \][/tex]
[tex]\[ 148 = 19k_1 + 123.5k_2 \][/tex]

### Solving the System of Equations:
We have the following system of linear equations:

1. [tex]\( 15k_1 + 24k_2 = 129 \)[/tex]
2. [tex]\( 19k_1 + 123.5k_2 = 148 \)[/tex]

We will solve these equations to find [tex]\(k_1\)[/tex] and [tex]\(k_2\)[/tex].

### Solution:
From the calculations, we find:
[tex]\[ k_1 = 8.86466165413534 \][/tex]
[tex]\[ k_2 = -0.16541353383458654 \][/tex]

### Interpretation of Constants:
1. [tex]\( k_1 = 8.86466165413534 \)[/tex]: This is the coefficient representing the contribution of the volume of water to the number of items produced.
2. [tex]\( k_2 = -0.16541353383458654 \)[/tex]: This is the coefficient representing the contribution of the product of the number of machines and the quantity of raw materials to the number of items produced.

Thus, the relationship for the number [tex]\(N\)[/tex] of items produced by the company can be expressed as:

[tex]\[ N = 8.86466165413534 v - 0.16541353383458654 (A \times Q) \][/tex]

This equation shows how the production of items [tex]\(N\)[/tex] is influenced by the volume of water [tex]\(v\)[/tex] and the product of the number of machines [tex]\(A\)[/tex] and the quantity of raw materials [tex]\(Q\)[/tex].