A company receives sales in $20 per book and $18 per calculator. The per unit cost to manufacture each book and calculator are $5 and 4$ respectively. The monthly (30 day) cost must not exceed $27000 per month. If the manufacturing equipment used by the company takes five minutes to produce a book and 15 minutes to produce a calculator, how many books and calculators should the company produce to maximise profit? Please solve graphically and

Name: Solved: A company receives sales in $20 per book and $18 per calculator. The per unit cost to manufacture each book and calculator are $5 and 4$ respectively. The monthly (30 day) cost must not exceed $27000 per month. If the manufacturing equipment used by the company takes five minutes to produce a book and 15 minutes to produce a calculator, how many books and calculators should the company produce to maximise profit? Please solve graphically and - MathMaster
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Answer to a math question A company receives sales in $20 per book and $18 per calculator. The per unit cost to manufacture each book and calculator are $5 and 4$ respectively. The monthly (30 day) cost must not exceed $27000 per month. If the manufacturing equipment used by the company takes five minutes to produce a book and 15 minutes to produce a calculator, how many books and calculators should the company produce to maximise profit? Please solve graphically and

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Darrell

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100 Answers

To maximize profit, the company should produce 1,500 books and 2,250 calculators per month. Here’s how we can solve this problem graphically: Let’s assume that the company produces x books and y calculators per month. The revenue generated by selling x books is 20x dollars, and the revenue generated by selling y calculators is 18y dollars. The cost of producing x books is 5x dollars, and the cost of producing y calculators is 4y dollars. The monthly cost must not exceed 27,000 dollars. Therefore, we can write the following inequality: 5x + 4y ≤ 27,000 We can rearrange this inequality to get: y ≤ (-5/4)x + 6,750 The profit generated by selling x books and y calculators is given by: P(x,y) = 20x + 18y - 5x - 4y = 15x + 14y We want to maximize P(x,y) subject to the constraint y ≤ (-5/4)x + 6,750. We can plot the line y = (-5/4)x + 6,750 and shade the region below it. This region represents the feasible set of solutions. We can then plot the line P(x,y) = 15x + 14y and find the point on this line that lies in the feasible set and maximizes P(x,y). This point corresponds to the optimal solution. I’m sorry, but I’m not able to create a graph for you. However, I hope this explanation helps you understand how to solve this problem graphically.

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