A company has three factories making cars. The factories use different processes and produce cars at different rates. Factory X makes 5 fullsize cars per day, 5 midsize cars per day, and 5 economy cars per day. Factory Y makes 2 fullsize cars per day, 3 midsize cars per day, and 4 economy cars per day. Factory Z makes 16 fullsize cars per day, 19 midsize cars per day, and 22 economy cars per day. The company wants to produce 51 fullsize cars, 59 midsize cars, and 67 economy cars. The question is: how many days should each factory run to do so? The linear algebra problem has infinitely many solutions. Not all of these are physically possible. We care about the physically possible solutions. The physically possible solutions are: Factory X can be run for − 2 t + 7 days, Factory Y can be run for − 3 t + 8 days, and Factory Z can be run for 1 t + 0 days where 0 ≤ t ≤ . You need to choose how to deploy your factories. Each factory costs the same amount per day to run. To minimize your costs, you should run Factory X days, Factory Y 0 days, and Factory Z days.
A company has three factories making cars. The factories use different processes and produce cars at different rates. Factory X makes 5 fullsize cars per day, 5 midsize cars per day, and 5 economy cars per day. Factory Y makes 2 fullsize cars per day, 3 midsize cars per day, and 4 economy cars per day. Factory Z makes 16 fullsize cars per day, 19 midsize cars per day, and 22 economy cars per day. The company wants to produce 51 fullsize cars, 59 midsize cars, and 67 economy cars. The question is: how many days should each factory run to do so? The
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