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A little bit of background - I am currently in high school learning system of equations and inequalities. We've done things like linear inequalities, inequalities with parabolas, and a combination of those points.

I have a homework problem that I am a bit stumped on. In the problem below, I've substituted different numbers in in hopes that I can solve the original problem myself. The question reads:

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 $20x + 9y \le 800$. I also know that $x\ge0$ and $y\ge0$. 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.

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If the solution should have $x\approx y$, why not set $x=y$ and see what happens? –  AMPerrine Nov 2 '11 at 0:27
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3 Answers 3

up vote 3 down vote accepted

The question is not well-defined. One trivial solution is $x = y = 0$, but that's obviously not what they meant. Perhaps what they really meant is "try to seat as many people, while trying to keep $x \approx y$". This is a two-objective optimization problem, so there may be several "optimal" solutions.

If $x = y$, then we have $29x \leq 800$, and so $x \leq 27$, and we seat $783$ people.

If $x = y + 1$, then we have $29x \leq 809$, and so $x \leq 27$, and we seat $774$ people.

If $x = y - 1$, then we have $29x \leq 793$, and so $x \leq 27$, 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: $$ 783, 792, 772, 781, 790, 799, 779, 788, 797, 777, 786, 795, \ldots $$ The "best" solutions, given as $(20x+9y,|x-y|)$, are therefore $$ (783,0), (792,1), (794,2), (799,5), (800,11). $$

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Thank you, I have added that they are trying to seat the maximum amount of people. It seems to me, like before, that simply a rounding down of $800/29$ and adding and subtracting from $y$ will give me the optimal answers. I am still a bit curious what this question actually has to do with inequalities since there seems to be one optimal answer that seats the most people with the closest equal numbers of circular and rectangular tables. –  DMan Nov 2 '11 at 0:53
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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 $27 \times 20 + 28 \times 9=792$. As the problem said approximately, I would propose this answer. No inequality here, though.

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Using this method, I have gotten the correct answer on real problem with the numbers that was included on the page. While this is entirely intuitive, I'm still kind of worried about the inequality section. While it does not explicitly ask for one, it is a question in that unit and also asks for 'reasons'. +1 though. –  DMan Nov 2 '11 at 0:34
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Not an answer, but I was curious to see what this problem looked like.

I wrote a couple simple R functions that plot the valid combinations of rectangular and circular tables under certain user-specified circumstances. I've added the code I used at the end of the response (do double-check my code for typos/mistakes, though).

First, we have the valid combinations of circular and rectangular tables (Blue corresponds to less than or equal to 800 people and red otherwise: enter image description here

Now, we're interested in when $x$ "close to" $y$. My first function plots the valid combinations such that $|x-y|\leq k$ where the user inputs $k$. For example, here's a plot of the valid combinations when $|x-y|\leq 5$:

enter image description here

Finally, we want to get the number of people seated close to 800. My second function plots the valid combinations such that $|x-y|\leq k$ and such that $20x+9y\geq m$ where the user inputs $k$ and $m$. Here's a plot of the valid combinations when $|x-y|\leq 5$ and $20x+9y\geq 785$ (the numbers represent the actual number seated with the given combination -- hard to see in the version I've written):

enter image description here

Again, I realize this doesn't answer the question but I think it helps to visualize the problem. And you can run my code, changing up the parameters.

R Code:

### first plot
x <- 0:40
y <- 0:88

plot(seq(0,40,length=50),seq(0,88,length=50),type="n",ylab="Circular   
     Tables", xlab="Rectangular Tables")
abline(v=(x), col="lightgray", lty="dotted")
abline(h=(y), col="lightgray", lty="dotted")

for(i in 1:length(x)){
  for(j in 1:length(y)){
    if(20*x[i]+9*y[j]<=800){
       points(x[i], y[j], col="Blue")
    }
    else{
       points(x[i], y[j], col="Red")
    }
  }
}

### function for second plot
under800almostequal <- function(k){
  plot(seq(0,40,length=50),seq(0,88,length=50),type="n",ylab="Circular
      Tables", xlab="Rectangular Tables")
  abline(v=(x), col="lightgray", lty="dotted")
  abline(h=(y), col="lightgray", lty="dotted")

  for(i in 1:length(x)){
    for(j in 1:length(y)){
      if(20*x[i]+9*y[j]<=800 && abs(x[i]-y[j])<=k){
         points(x[i], y[j], col="Blue")
      }
    }
  }
}

### function for third plot
under800almostequal2 <- function(k, m){
  plot(seq(0,40,length=50),seq(0,88,length=50),type="n",ylab="Circular 
      Tables", xlab="Rectangular Tables")
  abline(v=(x), col="lightgray", lty="dotted")
  abline(h=(y), col="lightgray", lty="dotted")

  for(i in 1:length(x)){
    for(j in 1:length(y)){
      if(20*x[i]+9*y[j]<=800 && abs(x[i]-y[j])<= k &&
         20*x[i]+9*y[j] >= m){
        text(x[i], y[i], 20*x[i]+9*y[j], cex=.85)
      }
    }
  }
}
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Thank you! This does indeed help me see the problem as the linear inequalities we have been solving. I can pretty intuitively see that at around 27, 27 (which is around the right answer), the $x$ and $y$ values are about the same and then go to the point that rests on the boundary line. +1 for including the graphs, and because I might pick up R now ;) –  DMan Nov 2 '11 at 2:20
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