In a general sense: Let a<b<c the sides of the rectangle. Then, the corresponding volumes of inscribed right cilinders would be:

pi*a^2/4*c

pi*b^2/4*a

pi*a^2/4*b

The last one is always the smallest volume. Now, between the 1st and the 2nd, itÂ´ll depend on whether ac<>b^2. In the particular case of the question U brought about, b=10, and b^2=100 is in effect larger than 8*12=96. Had the smallest side been 9 instead of 8, weÂ´d have had the other case: 10^2<9*12(=108).

Hope this helps.

bmwhype2 wrote:

a rectangular box has 12 x 10 x 8 dimensions. what is the largest possible volume of a right cylinder that is placed inside the box?

Answer:

V= (pi) (r^2) H

V= (5^2) pi x 8

V=200pi

the explanation was the radius of the cylinder must be equal to half of the smaller of the 2 dimensions of the box's bottom. the largest possible radius will yield the largest volume.

My question is..... why didnt we take half of 12 instead of 10 since we are looking for the largest radius?