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by vinviper1 » Mon Apr 07, 2008 11:02 am
rectangular dimensions are 6 by 8 by 10 inches. A cylindrical canister is to be placed insde the box so that it stands upgright whtn teh closed box rests on oen of its six faces. OF all such canisters that could be used, what is teh radius, in inches, of the one that has maximum volume?

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by xilef » Mon Apr 07, 2008 11:21 am
Well, you can eliminate C,D and E because the largest side of the cube is 10 by 8 so the largest cylinder you can have is with diameter of 8in. Basically, you can only have 2 diameters, 6in and 8in. The volume of a cylinder is h*area of the circle. You have 2 circle areas: 9pi and 16pi now you can go through the possible volumes to find the largest one:

9pi*10=90pi
9pi*8=72pi
16pi*6=96pi - winner, so the r that we want is 4.
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by vinviper1 » Tue Apr 08, 2008 7:20 am
Thanks. Now when they say 6 by 8 by 10 inches -- are they limiting it to a few being L and W, or are they all interchangable.

Like why cant 10 inches by the potential circumference aka 5 squared Pi = area == 25Pi times 8 for the height. I know I am missing something.

Thanks!
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by GMAT_crusher » Tue Apr 08, 2008 8:41 am
vinviper1 wrote:Thanks. Now when they say 6 by 8 by 10 inches -- are they limiting it to a few being L and W, or are they all interchangable.

Like why cant 10 inches by the potential circumference aka 5 squared Pi = area == 25Pi times 8 for the height. I know I am missing something.

Thanks!
10 x 8 base, gives max dia. 8 (radius 4mm); ht. 6mm
8 x 6 base, gives max dia 6 (radius 3mm); ht 10mm
6 x 10 base, gives max dia 6 (radius 3mm); ht 8mm

... I guess you get the picture now.
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by GMAT_crusher » Tue Apr 08, 2008 8:44 am
vinviper1 wrote:Thanks. Now when they say 6 by 8 by 10 inches -- are they limiting it to a few being L and W, or are they all interchangable.

Like why cant 10 inches by the potential circumference aka 5 squared Pi = area == 25Pi times 8 for the height. I know I am missing something.

Thanks!
10 x 8 base, gives max dia. 8 (radius 4mm); ht. 6mm
8 x 6 base, gives max dia 6 (radius 3mm); ht 10mm
6 x 10 base, gives max dia 6 (radius 3mm); ht 8mm

... I guess you get the picture now.
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by GMAT_crusher » Tue Apr 08, 2008 9:32 am
vinviper1 wrote:Thanks. Now when they say 6 by 8 by 10 inches -- are they limiting it to a few being L and W, or are they all interchangable.

Like why cant 10 inches by the potential circumference aka 5 squared Pi = area == 25Pi times 8 for the height. I know I am missing something.

Thanks!
10 x 8 base, gives max dia. 8 (radius 4mm); ht. 6mm
8 x 6 base, gives max dia 6 (radius 3mm); ht 10mm
6 x 10 base, gives max dia 6 (radius 3mm); ht 8mm

... I guess you get the picture now.
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by vinviper1 » Tue Apr 08, 2008 12:05 pm
Thanks,

What about 10 X 6 -- Max diameter of 10 height of 8?
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by xilef » Tue Apr 08, 2008 3:04 pm
vinviper1, maybe visual explanation will help. In a 10 by 6 rectangle the largest inscribed circle has a diameter of 6.
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by Stuart@KaplanGMAT » Tue Apr 08, 2008 4:25 pm
The diameter is limited by the shortest dimension of that side, not the longest.

For example, in a 10*6 side, the biggest possible diameter of an inscribed circle will be 6.

We want the cylinder with the greatest possible volume. The formula is:

volume = pi(r^2) * h

If we think about the dimensions, we see that since radius is squared and height is linear, increasing the radius will have a bigger impact than increasing the height. Therefore, we want to choose an orientation that maximizes radius.

To do so, we'll set the base of the cylinder on the 8*10 side, which will give us a diameter of 8 and radius of 4. Since the remaining dimension is 6, that will be our height.

So:

V = pi(r^2) * h = pi(16)(6) = 96pi.
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by vinviper1 » Thu Apr 10, 2008 12:52 pm
Ahhh...Thanks all for the extra help on this one.
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by yalephd2007 » Sun Apr 13, 2008 9:02 am
Smart solution! :D
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