David Allen
Posted on Tuesday, 14 January, 2003 - 11:37 pm:
Here is one that has me going ''WHAT??????''

Tennis balls used to be supplied in cardboard boxes holding two rows of three balls.

To maintain the balls in good condition, they are now supplied in closed airtight cylindrical tubes, made of metal or plastic: these tubes hold either three or four tennis balls.

In each case the balls just touch the adjacent balls and also the edges of their respective container: they are not distorted in any way.

Taking the radius of each ball to be r, and assuming that the balls fit neatly into their respective containers, answer the following.


    (a)

      (i) State the internal dimensions of the box and hence find its volume.
      (ii) State the internal dimensions of a tube made to contain three balls, and hence show that its volume is 6π r3 .
      (iii) Find the ratio of the volume of the tube to the volume of the box.
      (iv) In each case, what proportions of the available space is occupied by the balls? What conclusions can you draw?

    (b)

      (i) Calculate the surface area of the box. You can assume that the lid of the box has the same dimensions as the base.
      (ii) Calculate the surface area of the tube.
      (iii) Compare the amounts of material required to make one box or two tubes that will hold six tennis balls. What conclusions can you draw?

Maybe I'm just tired, as it is 11.35pm but I have tried several times to get my head round this and I can't. I think I need sleep, lol.

Anyone who can help, I need some guidance. Many thanks,

Dave
Hauke Worpel
Posted on Wednesday, 15 January, 2003 - 05:47 am:
Okay, I'll give it a shot.

(a)(i) The best way to do this is to visualize the balls lying flat on a table. Seen from above they'd look like this:

OO
OO
OO

Two rows of three. Now, the height of the box is 2r because the box is one ball high. The width is 4r (two balls) and the length is 6r (three balls).

2r x 4r x 6r = 48r3

(ii) This is easy. Three balls high, which you already know is 6r, times a circular base which should be the exact cross section of the ball. You should know how to find the area of a circle with radius r.

(iii) This is just division. You should be able to do this now that you know the volumes involved.

(iv) The volume of a sphere is 4pi r3 /3. Find the volume of six or four tennis balls or whatever and divide that by the volume of the container.

(b)(i) The easiest way to think about this is to realize that the box has six faces to it. You now know the dimensions of the box, so you should be able to get the side lengths of the faces. Draw a diagram of the box. It will help you immensely.

(ii) The cylinder has three bits to it. The base, the lid (which is the same as the base) and the cylindrical piece between. The lid and the base should present no difficulty. Imagine that you cut the curved face open and flatten it out; you get a rectangle. One side of the rectangle is the height of the cylinder, the other side you can work out because it was once the circumference of a circle.

(iii) Again, now that you have the numbers this will be easy.

I hope that helps.

HW
David Allen
Posted on Thursday, 16 January, 2003 - 05:53 pm:
Thanks, just needed my brain kick started. When I looked at it today it was so much easier, lol.

:-)

Dave