A room can have up to 800 people. There are two types of tables: one, rectangular, seating 20 people; another, circular, that seats 9. If the organizers want to have approximately the same amount of rectangular and circular tables to seat the maximum of people possible, what combination of tables can they use?Initially, I started off with a simple let statement, stating that x will represent rectangular tables, and y will represent circular tables. So my inequality so far would be something like 20 x + 9 y ≤ 800. I also know that x ≥ 0 and y ≥ 0. However, with this equation, I am not sure if I can actually find out an equal amount of rectangular and circular tables. I have also considered that this is a maximum/minimum problem that I have learned before, but the inequality I came up with is not a quadratic equation.The question is multiple choice, and, substituting the correct values in, I can see that it works, but I have no idea how they arrived at the numbers. It seems to me that the question can be solved without this guess-and-check, so I'm interested in what you have to say about this problem.

Question

A room can have up to 800 people. There are two types of tables: one, rectangular, seating 20 people; another, circular, that seats 9. If the organizers want to have approximately the same amount of rectangular and circular tables to seat the maximum of people possible, what combination of tables can they use?Initially, I started off with a simple let statement, stating that   x will represent rectangular tables, and   y will represent circular tables. So my inequality so far would be something like   20  x  +  9  y  ≤  800. I also know that   x  ≥  0 and   y  ≥  0. However, with this equation, I am not sure if I can actually find out an equal amount of rectangular and circular tables. I have also considered that this is a maximum/minimum problem that I have learned before, but the inequality I came up with is not a quadratic equation.The question is multiple choice, and, substituting the correct values in, I can see that it works, but I have no idea how they arrived at the numbers. It seems to me that the question can be solved without this guess-and-check, so I&#039;m interested in what you have to say about this problem.

Alexia Hart · Accepted Answer

The question is not well-defined. One trivial solution is   x  =  y  =  0, but that&#039;s obviously not what they meant. Perhaps what they really meant is &quot;try to seat as many people, while trying to keep   x  ≈  y&quot;. This is a two-objective optimization problem, so there may be several &quot;optimal&quot; solutions.If   x  =  y, then we have   29  x  ≤  800, and so   x  ≤  27, and we seat   783 people.If   x  =  y  +  1, then we have   29  x  ≤  809, and so   x  ≤  27, and we seat   774 people.If   x  =  y  −  1, then we have   29  x  ≤  793, and so   x  ≤  27, and we seat   792 people.For   x  =  y  +  2 and   x  =  y  −  2, we can seat   794 and   772 people, accordingly.For   x  =  y  +  3 and   x  =  y  −  3, we can seat   785 and   781 people, accordingly.For   x  =  y  +  4 and   x  =  y  −  4, we can seat   776 and   790 people, accordingly.For   x  =  y  +  5 and   x  =  y  −  5, we can seat   796 and   799 people, accordingly.The sequence for   x  =  y  +  d goes like  783  ,  774  ,  794  ,  785  ,  776  ,  796  ,  787  ,  778  ,  798  ,  789  ,  780  ,  800.Each time we either subtract   9 or add   20 (prove!). For   x  =  y  −  d, it goes the other way:  783  ,  792  ,  772  ,  781  ,  790  ,  799  ,  779  ,  788  ,  797  ,  777  ,  786  ,  795  ,  …The &quot;best&quot; solutions, given as   (  20  x  +  9  y  ,  |  x  −  y  |  ), are therefore  (  783  ,  0  )  ,  (  792  ,  1  )  ,  (  794  ,  2  )  ,  (  799  ,  5  )  ,  (  800  ,  11  )  .

gaiaecologicaq2 · Accepted Answer

If you have one table of each type, the pair seats   29 people.   27 of these pairs will seat   783 people. If we want to be as close to   800 as possible without going over, you could add one   9, leading to   27  ×  20  +  28  ×  9  =  792. As the problem said approximately, I would propose this answer. No inequality here, though.

A room can have up to 800 people. There are two types of tables: one, rectangular, seating 20 people

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