Logic

yovis

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I need help in a few problems from my homework.
Theotis has to load a truck with television sets. the cargo area of the truck is a rectangular block that measures 8ft by 2 ft by 11 ft. each television set measures 1 1/2 ft by 1 2/3ft by 1 1/3ft. What is the maximum number of tv sets that can be loaded into the truck?
 
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the cargo area of the truck is a rectangular block that measures 8 ft by 2 ft by 11 ft.


each television set measures 1+1/2 ft by 1+2/3 ft by 1+1/3 ft.

What is the maximum number of tv sets that can be loaded into the truck?



You need to divide the volume of the cargo area by the volume of one TV set.


Have you learned the volume formula for rectangular solids? We get volume by multiplying the measurements of the three dimensions together.

Volume = Length × Width × Height


Start by calculating the volume of the truck's cargo area. Its three dimensions are highlighted in red above. What volume do you get for the truck?


Next, calculate the volume of one TV set. Hint: Convert the mixed numbers (highlighted in blue above) to improper fractions first. What volume do you get for one TV?
 
I would point out that simply dividing the volume of the truck by the volume of a set will not necessarily answer the question. It will give an upper bound for the number of sets that you can fit, but since the dimensions of the truck are not evenly divisible by the set dimensions, fitting something approaching the theoretical maximum becomes a real challenge. IF you can demonstrate a way to arrange the boxes which achieves the maximum number possible, then you have a solution. If you cannot find a way to do it, it does not mean necessarily that there IS no way, but maybe just that you need to keep trying.

In general, packing problems like this are extremely difficult.
 


Is this the case at the pre-algebra level?

Maybe. The answer is certainly not 52, you can only put boxes one layer deep. 42 is one possibility but it is not obvious that it is the best you can do without some additional fiddling, and proving you have the best answer takes work.
 
Agreed -- I did not consider any wasted space, and it could very well be that this is intended (although I've seen a lot of poorly-designed "real-world" word problems posed).

Now I'm wondering whether those TV boxes are printed with "This Side UP!" warnings. ;)
 
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I would point out that simply dividing the volume of the truck by the volume of a set will not necessarily answer the question. It will give an upper bound for the number of sets that you can fit, but since the dimensions of the truck are not evenly divisible by the set dimensions, fitting something approaching the theoretical maximum becomes a real challenge. IF you can demonstrate a way to arrange the boxes which achieves the maximum number possible, then you have a solution. If you cannot find a way to do it, it does not mean necessarily that there IS no way, but maybe just that you need to keep trying.

In general, packing problems like this are extremely difficult.
Very true but the problem said maximum number of boxes that can be loaded on the truck. Not being able to evenly fit boxes would reduce, not increase the number of boxes you can fit in. Just dividing the volume of the truck by the volume of one box will give a quick maximum.
 
Maybe the truck is owned by Wal-Mart. Maybe the guys loading the televisions are Wal-Mart employees. They'll make 'em fit, using Wal-Mart logic.
 
I guess one way to solve may be to consider all possible fittings and find the maximum.

8 feet can be divide into 3 ways depending on how the TV (1.5x1.66x1.33) fits. Thus each dimension of TV gives us 8/1.5, 8/1.66 and 8/1.33. Since we can only take integer values, we get.

8--> 5, 4, 6
2--->1, 1, 1
11-->7, 6, 8

Now, the possible combinations of fitting the TVs are:

5x1x8, 5x1x6
4x1x8,4x1x7
6x1x7,6x1x7

Of these combinations, 42 seems like the maximum number of TVs that can be fit. This can be a reasonable answer as already mentioned by girodd.
 
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