16Oz Jeans has factories in Mydney and Selbourne. At the Mydney factory, fixed costs are $28000 per month and the cost of producing each pair of jeans is $30. At the Selbourne factory, fixed costs are $35200 per month and the cost of producing each pair of jeans is $24. During the next month Oz Jeans must manufacture 6000 pairs of jeans. Calculate the production order for each factory, if the total manufacturing costs for each factory are to be the same.
Q. 16Oz Jeans has factories in Mydney and Selbourne. At the Mydney factory, fixed costs are $28000 per month and the cost of producing each pair of jeans is $30. At the Selbourne factory, fixed costs are $35200 per month and the cost of producing each pair of jeans is $24. During the next month Oz Jeans must manufacture 6000 pairs of jeans. Calculate the production order for each factory, if the total manufacturing costs for each factory are to be the same.
Define Variables: Let's call the number of jeans produced in Mydney x and in Selbourne y. We know that x+y=6000 because the total number of jeans to be produced is 6000.
Calculate Total Costs: The total cost for Mydney is the fixed cost plus the variable cost of producing x jeans, so it's $28000+$30x. The total cost for Selbourne is the fixed cost plus the variable cost of producing y jeans, so it's $35200+$24y.
Set Cost Equations Equal: We want the costs to be the same, so we set the two expressions equal to each other:28000+30x=35200+24y
Solve for y: Now we solve for one of the variables. Let's solve for y in terms of x:30x−24y=35200−2800030x−24y=7200
Simplify Equation: Divide everything by $6 to simplify: \$\(5x - \$\(4\)y = \$\(1200\)
Eliminate Variable y: Now we remember that \(x + y = 6000\). Let's multiply this equation by \$(\$)\(4\) to help us eliminate y:\(\newline\)\$(\$)\(4\)x + \$(\$)\(4\)y = \$(\$)\(24000\)
Find Value of x: Add this to the \(5x - 4y = 1200\) equation:\(\newline\)\(5x - 4y + 4x + 4y = 1200 + 24000\)\(\newline\)\(9x = 25200\)
Substitute \(x\) to Find \(y\): Divide by \(\$9\) to find \(x\):\[x = \frac{\$25200}{\$9}\]\[x = 2800\]
Substitute \(x\) to Find \(y\): Divide by \(\$9\) to find \(x\): \(x = \frac{\$25200}{\$9}\) \(x = 2800\)Now we substitute \(x\) back into \(x + y = 6000\) to find \(y\): \(2800 + y = 6000\) \(y\)\(0\) \(y\)\(1\)
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