Answer :
Certainly! Let's thoroughly break down the problem step-by-step.
### Problem Statement:
The number of items, [tex]\( P \)[/tex], produced by a company:
1. Partly varies directly as the volume of water, [tex]\( v \)[/tex].
2. Partly varies jointly as the number of machines, [tex]\( A \)[/tex], and the quantity of raw materials, [tex]\( Q \)[/tex].
We have the following conditions:
- With 4 machines, 6 silos of raw materials, and 15 cubic meters of water, the company produced 129 boxes of items.
- With 4 machines, [tex]\( \sqrt{68} \)[/tex] silos of raw materials, and [tex]\( 19^3 \)[/tex] cubic meters of water, the company produced 148 boxes of items.
### Objective:
Find the proportionality constants [tex]\( k_1 \)[/tex] and [tex]\( k_2 \)[/tex].
### Mathematical Formulation:
Since the number of items produced, [tex]\( P \)[/tex], is partly directly proportional to the volume of water [tex]\( v \)[/tex], and partly jointly proportional to the number of machines [tex]\( A \)[/tex] and the quantity of raw materials [tex]\( Q \)[/tex], we can formulate the equation as:
[tex]\[ P = k_1 \cdot v + k_2 \cdot A \cdot Q \][/tex]
Where:
- [tex]\( k_1 \)[/tex] is the constant of proportionality for the volume of water.
- [tex]\( k_2 \)[/tex] is the constant of proportionality for the product of the number of machines and the quantity of raw materials.
### Step-by-Step Solution:
#### Step 1: Formulate the Equations
Using the given data, we form two equations based on the conditions given:
1. For the first condition:
[tex]\[ 129 = k_1 \cdot 15 + k_2 \cdot 4 \cdot 6 \][/tex]
Simplifying this:
[tex]\[ 129 = 15k_1 + 24k_2 \quad \text{(Equation 1)} \][/tex]
2. For the second condition:
[tex]\[ 148 = k_1 \cdot 19^3 + k_2 \cdot 4 \cdot \sqrt{68} \][/tex]
Calculating the specific values:
[tex]\[ 19^3 = 6859 \][/tex]
Thus:
[tex]\[ 148 = k_1 \cdot 6859 + k_2 \cdot 4 \cdot \sqrt{68} \][/tex]
Simplifying this:
[tex]\[ 148 = 6859k_1 + 4\sqrt{68}k_2 \quad \text{(Equation 2)} \][/tex]
#### Step 2: Solving the Simultaneous Equations
Now, we solve the simultaneous equations [tex]\( 15k_1 + 24k_2 = 129 \)[/tex] and [tex]\( 6859k_1 + 4\sqrt{68}k_2 = 148 \)[/tex].
#### Step 3: Extracting the Constants
By solving these equations, we find the values of [tex]\( k_1 \)[/tex] and [tex]\( k_2 \)[/tex]. The solutions are:
[tex]\[ k_1 = \frac{1011477}{47045456} - \frac{294197\sqrt{17}}{47045456} \][/tex]
[tex]\[ k_2 = \frac{1470985\sqrt{17}}{376363648} + \frac{2017897223}{376363648} \][/tex]
So, the constants are:
[tex]\[ k_1 = \frac{1011477}{47045456} - \frac{294197\sqrt{17}}{47045456} \][/tex]
[tex]\[ k_2 = \frac{1470985\sqrt{17}}{376363648} + \frac{2017897223}{376363648} \][/tex]
### Conclusion
The constants [tex]\( k_1 \)[/tex] and [tex]\( k_2 \)[/tex] that define the relationship between the number of items produced and the variables given are:
[tex]\[ k_1 = \frac{1011477}{47045456} - \frac{294197\sqrt{17}}{47045456} \][/tex]
[tex]\[ k_2 = \frac{1470985\sqrt{17}}{376363648} + \frac{2017897223}{376363648} \][/tex]
These constants can now be used in the original formula to predict the number of items produced for any given values of [tex]\( A \)[/tex], [tex]\( Q \)[/tex], and [tex]\( v \)[/tex].
### Problem Statement:
The number of items, [tex]\( P \)[/tex], produced by a company:
1. Partly varies directly as the volume of water, [tex]\( v \)[/tex].
2. Partly varies jointly as the number of machines, [tex]\( A \)[/tex], and the quantity of raw materials, [tex]\( Q \)[/tex].
We have the following conditions:
- With 4 machines, 6 silos of raw materials, and 15 cubic meters of water, the company produced 129 boxes of items.
- With 4 machines, [tex]\( \sqrt{68} \)[/tex] silos of raw materials, and [tex]\( 19^3 \)[/tex] cubic meters of water, the company produced 148 boxes of items.
### Objective:
Find the proportionality constants [tex]\( k_1 \)[/tex] and [tex]\( k_2 \)[/tex].
### Mathematical Formulation:
Since the number of items produced, [tex]\( P \)[/tex], is partly directly proportional to the volume of water [tex]\( v \)[/tex], and partly jointly proportional to the number of machines [tex]\( A \)[/tex] and the quantity of raw materials [tex]\( Q \)[/tex], we can formulate the equation as:
[tex]\[ P = k_1 \cdot v + k_2 \cdot A \cdot Q \][/tex]
Where:
- [tex]\( k_1 \)[/tex] is the constant of proportionality for the volume of water.
- [tex]\( k_2 \)[/tex] is the constant of proportionality for the product of the number of machines and the quantity of raw materials.
### Step-by-Step Solution:
#### Step 1: Formulate the Equations
Using the given data, we form two equations based on the conditions given:
1. For the first condition:
[tex]\[ 129 = k_1 \cdot 15 + k_2 \cdot 4 \cdot 6 \][/tex]
Simplifying this:
[tex]\[ 129 = 15k_1 + 24k_2 \quad \text{(Equation 1)} \][/tex]
2. For the second condition:
[tex]\[ 148 = k_1 \cdot 19^3 + k_2 \cdot 4 \cdot \sqrt{68} \][/tex]
Calculating the specific values:
[tex]\[ 19^3 = 6859 \][/tex]
Thus:
[tex]\[ 148 = k_1 \cdot 6859 + k_2 \cdot 4 \cdot \sqrt{68} \][/tex]
Simplifying this:
[tex]\[ 148 = 6859k_1 + 4\sqrt{68}k_2 \quad \text{(Equation 2)} \][/tex]
#### Step 2: Solving the Simultaneous Equations
Now, we solve the simultaneous equations [tex]\( 15k_1 + 24k_2 = 129 \)[/tex] and [tex]\( 6859k_1 + 4\sqrt{68}k_2 = 148 \)[/tex].
#### Step 3: Extracting the Constants
By solving these equations, we find the values of [tex]\( k_1 \)[/tex] and [tex]\( k_2 \)[/tex]. The solutions are:
[tex]\[ k_1 = \frac{1011477}{47045456} - \frac{294197\sqrt{17}}{47045456} \][/tex]
[tex]\[ k_2 = \frac{1470985\sqrt{17}}{376363648} + \frac{2017897223}{376363648} \][/tex]
So, the constants are:
[tex]\[ k_1 = \frac{1011477}{47045456} - \frac{294197\sqrt{17}}{47045456} \][/tex]
[tex]\[ k_2 = \frac{1470985\sqrt{17}}{376363648} + \frac{2017897223}{376363648} \][/tex]
### Conclusion
The constants [tex]\( k_1 \)[/tex] and [tex]\( k_2 \)[/tex] that define the relationship between the number of items produced and the variables given are:
[tex]\[ k_1 = \frac{1011477}{47045456} - \frac{294197\sqrt{17}}{47045456} \][/tex]
[tex]\[ k_2 = \frac{1470985\sqrt{17}}{376363648} + \frac{2017897223}{376363648} \][/tex]
These constants can now be used in the original formula to predict the number of items produced for any given values of [tex]\( A \)[/tex], [tex]\( Q \)[/tex], and [tex]\( v \)[/tex].
