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.

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!

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.

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.

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.

Thanks,

What about 10 X 6 -- Max diameter of 10 height of 8?

vinviper1, maybe visual explanation will help. In a 10 by 6 rectangle the largest inscribed circle has a diameter of 6.

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.

Ahhh...Thanks all for the extra help on this one.

Smart solution!