A paint supplier has two machines that produce both indoor and outdoor paint. To meet one of its contractual obligations, the company must produce at least 60 gallons of indoor paint and 100 gallons of outdoor paint per day.

- Machine A makes 3 gallons of indoor paint and 10 gallons of outdoor paint per hour.
- Machine B makes 4 gallons of indoor paint and 5 gallons of outdoor paint per hour.
- It costs $28 per hour to run Machine A and $33 per hour to run Machine B.

How many hours should each machine be operated to minimize the cost, and what is the minimum cost?

Let:
- [tex] x [/tex] = number of hours Machine A is operated
- [tex] y [/tex] = number of hours Machine B is operated

This is a Linear Programming problem. Form inequalities using the information given, then draw them on a graph to identify the feasible region. Minimize cost within that region to find the optimal number of hours each machine should run.

Explanation:

This is a classic problem of the branch of mathematics known as Linear Programming. Since machines A and B produce both indoor and outdoor paint, the minimum number of hours they should be run to produce at least 60 gallons of indoor paint and 100 gallons of outdoor paint could be found by using linear inequalities and then finding the optimum solution that would minimize the cost of operating the machines.

Starting with the indoor paint, it translates to the inequality 3x + 4y >= 60. For the outdoor paint, this translates to the inequality 10x + 5y >= 100. Then you need to consider the total cost which is given by cost = 28x + 33y. By combining these inequalities and plotting them on a graph, you'll be able to identify the feasible region. Minimizing the costs within that region allows you to identify the optimum working hours for each machine.

This problem would typically be solved using a graphical method or simplex algorithm. Take note that the solution to this problem would vary depending upon the exact numbers. But the approach explained here will provide you a suitable starting point.

Learn more about Linear Programming here:

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