To solve the given problem, we need to find the relationship between the number of items produced, the volume of water, the number of machines, and the quantity of raw materials.
Let's denote:
- [tex]\( P \)[/tex] as the number of items produced,
- [tex]\( v \)[/tex] as the volume of water (in cubic meters),
- [tex]\( A \)[/tex] as the number of machines,
- [tex]\( Q \)[/tex] as the quantity of raw materials.
From the problem, we know that the number of items produced [tex]\( P \)[/tex] 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]. We can express this relationship with an equation:
[tex]\[ P = k_1 v + k_2 A Q \][/tex]
Where [tex]\( k_1 \)[/tex] and [tex]\( k_2 \)[/tex] are constants that we need to determine.
We are given two scenarios to create our equations:
Scenario 1:
- 4 machines,
- 6 silos of raw materials,
- 15 cubic meters of water,
- 129 items produced.
Plug these values into our equation:
[tex]\[ 129 = k_1 \cdot 15 + k_2 \cdot 4 \cdot 6 \][/tex]
Simplifying:
[tex]\[ 129 = 15k_1 + 24k_2 \quad \text{(Equation 1)} \][/tex]
Scenario 2:
- 4 machines,
- 6 silos of raw materials,
- 19 cubic meters of water,
- 148 items produced.
Plug these values into our equation:
[tex]\[ 148 = k_1 \cdot 19 + k_2 \cdot 4 \cdot 6 \][/tex]
Simplifying:
[tex]\[ 148 = 19k_1 + 24k_2 \quad \text{(Equation 2)} \][/tex]
We now have a system of linear equations:
1. [tex]\( 15k_1 + 24k_2 = 129 \)[/tex]
2. [tex]\( 19k_1 + 24k_2 = 148 \)[/tex]
To solve this system, we can subtract Equation 1 from Equation 2 to eliminate [tex]\( k_2 \)[/tex]:
[tex]\[ (19k_1 + 24k_2) - (15k_1 + 24k_2) = 148 - 129 \][/tex]
[tex]\[ 4k_1 = 19 \][/tex]
[tex]\[ k_1 = \frac{19}{4} \][/tex]
Now that we have [tex]\( k_1 \)[/tex], we substitute it back into Equation 1 to find [tex]\( k_2 \)[/tex]:
[tex]\[ 15 \left(\frac{19}{4}\right) + 24k_2 = 129 \][/tex]
[tex]\[ \frac{285}{4} + 24k_2 = 129 \][/tex]
[tex]\[ 24k_2 = 129 - \frac{285}{4} \][/tex]
[tex]\[ 24k_2 = \frac{516}{4} - \frac{285}{4} \][/tex]
[tex]\[ 24k_2 = \frac{231}{4} \][/tex]
[tex]\[ k_2 = \frac{231}{4 \times 24} \][/tex]
[tex]\[ k_2 = \frac{231}{96} = \frac{77}{32} \][/tex]
Thus, the constants are:
- [tex]\( k_1 = \frac{19}{4} \)[/tex]
- [tex]\( k_2 = \frac{77}{32} \)[/tex]
Therefore, the relationship between the number of items produced, the volume of water, the number of machines, and the quantity of raw materials is given by:
[tex]\[ P = \frac{19}{4} v + \frac{77}{32} A Q \][/tex]
