IDNLearn.com: Where your questions meet expert advice and community support. Join our interactive Q&A community and access a wealth of reliable answers to your most pressing questions.
Sagot :
Sure, let's solve the problem step-by-step.
You want to determine the minimum cost to run machine [tex]\( X \)[/tex] and machine [tex]\( Y \)[/tex] to produce the required number of units of products [tex]\( A \)[/tex] and [tex]\( B \)[/tex].
Given:
- Machine [tex]\( X \)[/tex] produces 4 units of [tex]\( A \)[/tex] and 5 units of [tex]\( B \)[/tex] per hour.
- Machine [tex]\( Y \)[/tex] produces 3 units of [tex]\( A \)[/tex] and 10 units of [tex]\( B \)[/tex] per hour.
- Cost to run machine [tex]\( X \)[/tex] is [tex]$22 per hour. - Cost to run machine \( Y \) is $[/tex]25 per hour.
Let:
- [tex]\( x \)[/tex] be the number of hours machine [tex]\( X \)[/tex] runs.
- [tex]\( y \)[/tex] be the number of hours machine [tex]\( Y \)[/tex] runs.
The objective function [tex]\( C \)[/tex] representing the total cost is:
[tex]\[ C = 22x + 25y \][/tex]
Units produced per hour:
1. For [tex]\( A \)[/tex]:
[tex]\[ 4x + 3y \][/tex]
2. For [tex]\( B \)[/tex]:
[tex]\[ 5x + 10y \][/tex]
To determine the exact solution, we need the constraints related to required units of [tex]\( A \)[/tex] and [tex]\( B \)[/tex]. Assuming [tex]\( a \)[/tex] units of [tex]\( A \)[/tex] and [tex]\( b \)[/tex] units of [tex]\( B \)[/tex] are required, we have the following constraints:
1. [tex]\( 4x + 3y \geq a \)[/tex]
2. [tex]\( 5x + 10y \geq b \)[/tex]
3. [tex]\( x, y \geq 0 \)[/tex]
For simplicity, let's solve a hypothetical example where we need:
- [tex]\( a = 24 \)[/tex] units of [tex]\( A \)[/tex]
- [tex]\( b = 50 \)[/tex] units of [tex]\( B \)[/tex]
Then the constraints become:
1. [tex]\( 4x + 3y \geq 24 \)[/tex]
2. [tex]\( 5x + 10y \geq 50 \)[/tex]
Let's solve the system of inequalities:
### Step 1: Express [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex] from the second inequality:
[tex]\[ 5x + 10y \geq 50 \][/tex]
[tex]\[ 5x \geq 50 - 10y \][/tex]
[tex]\[ x \geq 10 - 2y \][/tex]
### Step 2: Substitute [tex]\( x \)[/tex] into the first inequality:
[tex]\[ 4(10 - 2y) + 3y \geq 24 \][/tex]
[tex]\[ 40 - 8y + 3y \geq 24 \][/tex]
[tex]\[ 40 - 5y \geq 24 \][/tex]
[tex]\[ -5y \geq -16 \][/tex]
[tex]\[ y \leq \frac{16}{5} \][/tex]
[tex]\[ y \leq 3.2 \][/tex]
So, our constraints are [tex]\( x \geq 10 - 2y \)[/tex] and [tex]\( y \leq 3.2 \)[/tex]. Let's substitute [tex]\( y = 3.2 \)[/tex]:
### Step 3: Find value of [tex]\( x \)[/tex]:
[tex]\[ x \geq 10 - 2(3.2) \][/tex]
[tex]\[ x \geq 10 - 6.4 \][/tex]
[tex]\[ x \geq 3.6 \][/tex]
### Step 4: To find minimum cost, let's find cost at [tex]\( y = 3.2, x = 3.6 \)[/tex]:
[tex]\[ C = 22x + 25y \][/tex]
[tex]\[ C = 22(3.6) + 25(3.2) \][/tex]
[tex]\[ C = 79.2 + 80 \][/tex]
[tex]\[ C = 159.2 \][/tex]
Therefore, the minimum cost to run the machines while meeting the production goals is:
[tex]\[ \$159.20 \][/tex]
This occurs when:
[tex]\[ x = 3.6 \text{ hours} \][/tex]
[tex]\[ y = 3.2 \text{ hours} \][/tex]
You want to determine the minimum cost to run machine [tex]\( X \)[/tex] and machine [tex]\( Y \)[/tex] to produce the required number of units of products [tex]\( A \)[/tex] and [tex]\( B \)[/tex].
Given:
- Machine [tex]\( X \)[/tex] produces 4 units of [tex]\( A \)[/tex] and 5 units of [tex]\( B \)[/tex] per hour.
- Machine [tex]\( Y \)[/tex] produces 3 units of [tex]\( A \)[/tex] and 10 units of [tex]\( B \)[/tex] per hour.
- Cost to run machine [tex]\( X \)[/tex] is [tex]$22 per hour. - Cost to run machine \( Y \) is $[/tex]25 per hour.
Let:
- [tex]\( x \)[/tex] be the number of hours machine [tex]\( X \)[/tex] runs.
- [tex]\( y \)[/tex] be the number of hours machine [tex]\( Y \)[/tex] runs.
The objective function [tex]\( C \)[/tex] representing the total cost is:
[tex]\[ C = 22x + 25y \][/tex]
Units produced per hour:
1. For [tex]\( A \)[/tex]:
[tex]\[ 4x + 3y \][/tex]
2. For [tex]\( B \)[/tex]:
[tex]\[ 5x + 10y \][/tex]
To determine the exact solution, we need the constraints related to required units of [tex]\( A \)[/tex] and [tex]\( B \)[/tex]. Assuming [tex]\( a \)[/tex] units of [tex]\( A \)[/tex] and [tex]\( b \)[/tex] units of [tex]\( B \)[/tex] are required, we have the following constraints:
1. [tex]\( 4x + 3y \geq a \)[/tex]
2. [tex]\( 5x + 10y \geq b \)[/tex]
3. [tex]\( x, y \geq 0 \)[/tex]
For simplicity, let's solve a hypothetical example where we need:
- [tex]\( a = 24 \)[/tex] units of [tex]\( A \)[/tex]
- [tex]\( b = 50 \)[/tex] units of [tex]\( B \)[/tex]
Then the constraints become:
1. [tex]\( 4x + 3y \geq 24 \)[/tex]
2. [tex]\( 5x + 10y \geq 50 \)[/tex]
Let's solve the system of inequalities:
### Step 1: Express [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex] from the second inequality:
[tex]\[ 5x + 10y \geq 50 \][/tex]
[tex]\[ 5x \geq 50 - 10y \][/tex]
[tex]\[ x \geq 10 - 2y \][/tex]
### Step 2: Substitute [tex]\( x \)[/tex] into the first inequality:
[tex]\[ 4(10 - 2y) + 3y \geq 24 \][/tex]
[tex]\[ 40 - 8y + 3y \geq 24 \][/tex]
[tex]\[ 40 - 5y \geq 24 \][/tex]
[tex]\[ -5y \geq -16 \][/tex]
[tex]\[ y \leq \frac{16}{5} \][/tex]
[tex]\[ y \leq 3.2 \][/tex]
So, our constraints are [tex]\( x \geq 10 - 2y \)[/tex] and [tex]\( y \leq 3.2 \)[/tex]. Let's substitute [tex]\( y = 3.2 \)[/tex]:
### Step 3: Find value of [tex]\( x \)[/tex]:
[tex]\[ x \geq 10 - 2(3.2) \][/tex]
[tex]\[ x \geq 10 - 6.4 \][/tex]
[tex]\[ x \geq 3.6 \][/tex]
### Step 4: To find minimum cost, let's find cost at [tex]\( y = 3.2, x = 3.6 \)[/tex]:
[tex]\[ C = 22x + 25y \][/tex]
[tex]\[ C = 22(3.6) + 25(3.2) \][/tex]
[tex]\[ C = 79.2 + 80 \][/tex]
[tex]\[ C = 159.2 \][/tex]
Therefore, the minimum cost to run the machines while meeting the production goals is:
[tex]\[ \$159.20 \][/tex]
This occurs when:
[tex]\[ x = 3.6 \text{ hours} \][/tex]
[tex]\[ y = 3.2 \text{ hours} \][/tex]
We greatly appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. Find reliable answers at IDNLearn.com. Thanks for stopping by, and come back for more trustworthy solutions.