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Here is our GeoGebra solution to example 1 except that now we have added sliders for
the total cost allowed and for the total floor space (footprint) available. What we are going
to do is move these sliders to make a new but similar problem for our students to practice
or for an exam question. In the second mathcast on this page, we will explain how to make
these sliders. But right now we are just going to explore with these sliders. The sliders
have 2 good points. Number 1 is that we can see if the problem is valid at all. Number
2 is that if we are working with something like in this problem where we needed the number
of filing cabinets, you must make sure – at minimum – that the answer is on a grid
point. Here the answer was (8,3) so 8 filing cabinets of type A and 3 filing cabinets of
type B. It is also good if the other intersection points are grid points.
Okay. So let us move our sliders here and let us see what happens. If we move ct down
towards 100. At 120, we are losing our problem. We don’t even have a good region. Anyway
we will keep going to 100. Then we will see what happens with the other slider. Do we
want to move this way? No, that makes it worse. So we want to move left. We see that E is
lying on a grid point and A is okay, but D does not look good. So this is a reasonable
problem, but not a good problem. Let us move ft again towards that grid point there. And
now we have good grid points for all the points, but most importantly for E, A and D.
Okay. Here we had just changed the totals. What would happen if we changed the parameters
in the problem. There are 6 parameters. Here we put in all 6 parameters as sliders. Again
we will show how to do this in the second mathcast. But 6 parameters is a lot of parameters
to be moving around. One of the key things is that the total number should be divisible
by both of these numbers (the coefficients). So, let us leave this as 10 and let us say
we make this 25. 140 is not divisible by 25 evenly, so let us say we make this 150. And
now we see that the points A and B are lying on grid points. We also see grid points here
and here on the line. So we will start to move these around. Let us make this one 1.
(Of course I do know a good answer before I started this mathcast.) And now let us move
this to a whole number. 7 let’s say. Then this cannot be a 0.8 for sure. It might be
able to be 0.7. And we see that we are not getting a good point for E. So let us move
it to 0.6 (it is not divisible). How about 0.5. The number 0.5 gives me all grid points.
(Like I said, I knew this answer ahead of time.)
You have to try moving the slider around and see what you get. Again, this is a good problem
for students to try. They get to find grid points for A, E and D and see how difficult
it is to actually make up a good problem and what a good problem looks like.