60 boxes, 11-by-6-by-4, must fit inside square-based box

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leish_peep

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Oct 11, 2008
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60 fruit drink containers size: 11cm x 6cm x 4cm
I need to fit all 60 into a carton that is square based.
it needs to be the smallest possible surface area.

Thanks a million... i dont mean to be pushy but asap would be absolutely fabulous :) :D :)
 
Re: Surface Area

"Minimum surface area." --- Is this of the entire carton, bottom, sides and top? bottom and sides? bottom only?
 
Re: Surface Area

this is minimum surface area of the whole thing including lid!
bottom, four sides, top....

THANKS SOO MUCH
 
Re: Surface Area

leish_peep said:
60 fruit drink containers size: 11cm x 6cm x 4cm
I need to fit all 60 into a carton that is square based.
it needs to be the smallest possible surface area.

Thanks a million... i dont mean to be pushy but asap would be absolutely fabulous :) :D :)
Click to expand...

Please show us your work/thoughts - indicating exactly where you are stuck - so that we know where to begin to help you.
 
Re: Surface Area

ahh well i have no workings cause i dunt get what to do....

all someone told me was that it had to multiples of 11, 6 and 4

im struggling
 
Re: Surface Area

60 fruit drink containers size: 11cm x 6cm x 4cm
I need to fit all 60 into a carton that is square based.
it needs to be the smallest possible surface area.

Suppose you want the base of the containers to be 6 cm x 4 cm and the height 11 cm.
We can name the length of one side of the base of the carton x cm. and the length of the other side of the carton y cm.
Since the base is square we know that (1) x=y.
The number of containers on the x side of the base is x/4.
The number of containers on the y side of the base is y/6.
(2) \(\displaystyle \frac{x}{4}\times \frac{y}{6}=60\)

Solve (1) and (2) for x or y to give you the length of the sides of the base of the carton.

You already know that the height of the carton is 11 cm. The surface are of the carton is 2xy + 2(11)x + 2(11)y So you can figure the total surface area for this configuration.

Next, you can find out the surface area of the carton if you have the bottom of the containers being 4 cm X 11 cm and the height 6 cm. Then do it if the base of the containers is 6 X 11 and the height 4. Finally, compare the three total surface areas to see which is the minimum.

Of course, my approach only covers the possibility that there is only one layer of containers. Possibly there could be two or more layers. You cold spend a lot of time on this. Maybe there is a better way.
 
Re: Surface Area

can you please tell us - how are you supposed to prove if some configuration you might choose is minimum or not?

Are you allowed to have gaps - that is unfilled space?

suppose you have a 24 x 24 square base.

Suppose you laid the carton so that - 11 by 4 is on the base and 6 cm height.

We can have 12 boxes in the base (with a 2* 24 space unoccupied) - 30 cm height

the area of the lid and the base is 2*24*24 and the sides are 4*24*30 = 4032 sq.cm

Is that minimum - you tell me!!!
 
Re: Surface Area

60 fruit drink containers size: 11cm x 6cm x 4cm
I need to fit all 60 into a carton that is square based.
it needs to be the smallest possible surface area.

Thanks a million... i dont mean to be pushy but asap would be absolutely fabulous

Considering all rectangles with the same perimeter, the square encloses the greatest area.
Proof: Consider a square of dimensions x by x, the area of which is x^2. Adjusting the dimensions by adding a to one side and subtracting a from the other side results in an area of (x + a)(x - a) = x^2 - a^2. Thus, however small the dimension "a" is, the area of the modified rectangle is always less than the square of area x^2.

*--Considering all rectangles with the same area, the square results in the smallest perimeter for a given area.

*--Considering all 3 sided boxes, the one with all 3 sides equal (a cube) has the minimum surface area.

Each container has 11x6x4 = 264 cub.cm.
The total volume of the box must be 264x60 = 15,840 cub.cm.
Therefore, the box of minimum surface are would be (15,840)^(1/3) or 25.114 x 25.114 x 25.114 cm.

A box with sides closest to 25.114cm would do the job.

How about a box 22 x 24 x 24 holding 2x6x4 = 48. Oops, short 12.

Perhaps you get the idea now. See what you can come up with.
 
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