Answer :
To address the given problem, where the production of items is partly dependent on the volume of water and partly jointly dependent on the number of machines and the quantity of raw material, we can model the situation using a linear equation involving these variables.
Let's break down the information and the steps to find the solution.
### Variables and Initial Information
1. Define the Variables:
- Let [tex]\(V\)[/tex] be the volume of water.
- Let [tex]\(A\)[/tex] be the number of machines.
- Let [tex]\(Q\)[/tex] be the quantity of raw material.
- Let [tex]\(P\)[/tex] be the number of items produced.
2. First Scenario:
- Number of machines ([tex]\(A\)[/tex]) = 4
- Quantity of raw material ([tex]\(Q\)[/tex]) = 6 silos
- Volume of water ([tex]\(V\)[/tex]) = 15 cubic meters
- Production ([tex]\(P\)[/tex]) = 129 boxes
3. Second Scenario:
- Number of machines ([tex]\(A\)[/tex]) = 4
- Quantity of raw material ([tex]\(Q\)[/tex]) = [tex]\(6/2 = 3\)[/tex] silos
- Volume of water ([tex]\(V\)[/tex]) = 19 cubic meters
- Production ([tex]\(P\)[/tex]) = 148 boxes
### Formulating the Equations
We know that the production [tex]\(P\)[/tex] partly depends on the volume of water [tex]\(V\)[/tex] and partly jointly on the number of machines [tex]\(A\)[/tex] and the quantity of raw material [tex]\(Q\)[/tex]. We can express this relationship mathematically as:
[tex]\[ P = k_1 \cdot V + k_2 \cdot A \cdot Q \][/tex]
Where [tex]\(k_1\)[/tex] and [tex]\(k_2\)[/tex] are constants that we need to determine.
Using the given scenarios, we set up the following equations:
1. For the first set of values:
[tex]\[ 129 = k_1 \cdot 15 + k_2 \cdot 4 \cdot 6 \][/tex]
2. For the second set of values:
[tex]\[ 148 = k_1 \cdot 19 + k_2 \cdot 4 \cdot 3 \][/tex]
### Solving the Equations
Now, solve these equations to find [tex]\(k_1\)[/tex] and [tex]\(k_2\)[/tex]:
#### Equation 1:
[tex]\[ 129 = 15k_1 + 24k_2 \][/tex] [tex]\(\rightarrow\)[/tex] (i)
#### Equation 2:
[tex]\[ 148 = 19k_1 + 12k_2 \][/tex] [tex]\(\rightarrow\)[/tex] (ii)
We can use the method of solving simultaneous equations.
First, we multiply equation (ii) by 2 to facilitate elimination:
[tex]\[ 296 = 38k_1 + 24k_2 \][/tex] [tex]\(\rightarrow\)[/tex] (iii)
Next, we subtract equation (i) from equation (iii):
[tex]\[ 296 - 129 = 38k_1 + 24k_2 - 15k_1 - 24k_2 \][/tex]
[tex]\[ 167 = 23k_1 \][/tex]
[tex]\[ k_1 = \frac{167}{23} = 7.26086956521739 \][/tex]
Now, substitute [tex]\(k_1\)[/tex] back into equation (i):
[tex]\[ 129 = 15(7.26086956521739) + 24k_2 \][/tex]
[tex]\[ 129 = 108.91304347826 + 24k_2 \][/tex]
[tex]\[ 129 - 108.91304347826 = 24k_2 \][/tex]
[tex]\[ 20.08695652174 = 24k_2 \][/tex]
[tex]\[ k_2 = \frac{20.08695652174}{24} = 0.83695652173913 \][/tex]
### Verification
To make sure these values are correct, let's verify by substituting [tex]\(k_1\)[/tex] and [tex]\(k_2\)[/tex] back into the original equations:
#### Verify with original equation (i):
[tex]\[ 129 = 15(7.26086956521739) + 24(0.83695652173913) \][/tex]
[tex]\[ 129 = 108.91304347826 + 20.08695652174 \][/tex]
[tex]\[ 129 = 129 \][/tex] (True)
#### Verify with original equation (ii):
[tex]\[ 148 = 19(7.26086956521739) + 12(0.83695652173913) \][/tex]
[tex]\[ 148 = 137.95652173913 + 10.04347826087 \][/tex]
[tex]\[ 148 = 148 \][/tex] (True)
Thus, the constants [tex]\(k_1\)[/tex] and [tex]\(k_2\)[/tex] are:
[tex]\[ k_1 = 7.26086956521739 \][/tex]
[tex]\[ k_2 = 0.83695652173913 \][/tex]
Hence, the solution to the problem is:
[tex]\[ k_1 \approx 7.261 \][/tex]
[tex]\[ k_2 \approx 0.837 \][/tex]
These constants accurately describe the relationship between the production of items, the volume of water, the number of machines, and the quantity of raw material for the given scenarios.
Let's break down the information and the steps to find the solution.
### Variables and Initial Information
1. Define the Variables:
- Let [tex]\(V\)[/tex] be the volume of water.
- Let [tex]\(A\)[/tex] be the number of machines.
- Let [tex]\(Q\)[/tex] be the quantity of raw material.
- Let [tex]\(P\)[/tex] be the number of items produced.
2. First Scenario:
- Number of machines ([tex]\(A\)[/tex]) = 4
- Quantity of raw material ([tex]\(Q\)[/tex]) = 6 silos
- Volume of water ([tex]\(V\)[/tex]) = 15 cubic meters
- Production ([tex]\(P\)[/tex]) = 129 boxes
3. Second Scenario:
- Number of machines ([tex]\(A\)[/tex]) = 4
- Quantity of raw material ([tex]\(Q\)[/tex]) = [tex]\(6/2 = 3\)[/tex] silos
- Volume of water ([tex]\(V\)[/tex]) = 19 cubic meters
- Production ([tex]\(P\)[/tex]) = 148 boxes
### Formulating the Equations
We know that the production [tex]\(P\)[/tex] partly depends on the volume of water [tex]\(V\)[/tex] and partly jointly on the number of machines [tex]\(A\)[/tex] and the quantity of raw material [tex]\(Q\)[/tex]. We can express this relationship mathematically as:
[tex]\[ P = k_1 \cdot V + k_2 \cdot A \cdot Q \][/tex]
Where [tex]\(k_1\)[/tex] and [tex]\(k_2\)[/tex] are constants that we need to determine.
Using the given scenarios, we set up the following equations:
1. For the first set of values:
[tex]\[ 129 = k_1 \cdot 15 + k_2 \cdot 4 \cdot 6 \][/tex]
2. For the second set of values:
[tex]\[ 148 = k_1 \cdot 19 + k_2 \cdot 4 \cdot 3 \][/tex]
### Solving the Equations
Now, solve these equations to find [tex]\(k_1\)[/tex] and [tex]\(k_2\)[/tex]:
#### Equation 1:
[tex]\[ 129 = 15k_1 + 24k_2 \][/tex] [tex]\(\rightarrow\)[/tex] (i)
#### Equation 2:
[tex]\[ 148 = 19k_1 + 12k_2 \][/tex] [tex]\(\rightarrow\)[/tex] (ii)
We can use the method of solving simultaneous equations.
First, we multiply equation (ii) by 2 to facilitate elimination:
[tex]\[ 296 = 38k_1 + 24k_2 \][/tex] [tex]\(\rightarrow\)[/tex] (iii)
Next, we subtract equation (i) from equation (iii):
[tex]\[ 296 - 129 = 38k_1 + 24k_2 - 15k_1 - 24k_2 \][/tex]
[tex]\[ 167 = 23k_1 \][/tex]
[tex]\[ k_1 = \frac{167}{23} = 7.26086956521739 \][/tex]
Now, substitute [tex]\(k_1\)[/tex] back into equation (i):
[tex]\[ 129 = 15(7.26086956521739) + 24k_2 \][/tex]
[tex]\[ 129 = 108.91304347826 + 24k_2 \][/tex]
[tex]\[ 129 - 108.91304347826 = 24k_2 \][/tex]
[tex]\[ 20.08695652174 = 24k_2 \][/tex]
[tex]\[ k_2 = \frac{20.08695652174}{24} = 0.83695652173913 \][/tex]
### Verification
To make sure these values are correct, let's verify by substituting [tex]\(k_1\)[/tex] and [tex]\(k_2\)[/tex] back into the original equations:
#### Verify with original equation (i):
[tex]\[ 129 = 15(7.26086956521739) + 24(0.83695652173913) \][/tex]
[tex]\[ 129 = 108.91304347826 + 20.08695652174 \][/tex]
[tex]\[ 129 = 129 \][/tex] (True)
#### Verify with original equation (ii):
[tex]\[ 148 = 19(7.26086956521739) + 12(0.83695652173913) \][/tex]
[tex]\[ 148 = 137.95652173913 + 10.04347826087 \][/tex]
[tex]\[ 148 = 148 \][/tex] (True)
Thus, the constants [tex]\(k_1\)[/tex] and [tex]\(k_2\)[/tex] are:
[tex]\[ k_1 = 7.26086956521739 \][/tex]
[tex]\[ k_2 = 0.83695652173913 \][/tex]
Hence, the solution to the problem is:
[tex]\[ k_1 \approx 7.261 \][/tex]
[tex]\[ k_2 \approx 0.837 \][/tex]
These constants accurately describe the relationship between the production of items, the volume of water, the number of machines, and the quantity of raw material for the given scenarios.
