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a rectangular box has 12 x 10 x 8 dimensions. what is the [#permalink]
18 Apr 2007, 15:50

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?

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?

The 3 surfaces are: 12*10,12*8,8*10

AMong the three 12*10 has the largest surface area. It can fit a cylinder of maximum diameter 10. Hence radius =5

V = pi*r^2*h= 25*9pi= 200pi

You can verify with other options:
For 12*8: r = 4, V=160pi
For 8*10:r =4,V=192pi

look in the diagram showing the box - the blue circle has diameter of 10 and the yellow circle a diameter of 12 - you will have to compromise and "lose" space in order to squeeze in the blue circle but you can't squeeze in the yellow one.

if you will pick 6 as the radius of the cylinder, the cylinder will not fit into the box (as shown in the diagram).

Re: volume question [#permalink]
01 May 2007, 05:41

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?

You have to take in consideration the radius and the hight of the cylinder. Then 6 is not the largest radius that gives the largest volume.

Re: volume question [#permalink]
01 May 2007, 08:07

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?

look in the diagram showing the box - the blue circle has diameter of 10 and the yellow circle a diameter of 12 - you will have to compromise and "lose" space in order to squeeze in the blue circle but you can't squeeze in the yellow one.

still confused. how exactly do we compromise (in numbers please )?

Re: volume question [#permalink]
02 Jun 2007, 03:01

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?

you can't take 12 as the cylinder will not fit into box then.

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

all right. i finally figured it out after some basic reasoning. a 12 inch ball cannot fit into a 12 inch box without some slack space.

therefore, we take the two largest dimensions as the base and then take the smaller of the base dimensions as a radius. thereby leaving the shortest dimension of the three as the height.

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?

If the base is 10 * 8, then volume is pi(4^2)(12) = 192pi
If the base is 10 * 12, then volume is pi(5^2)(8) = 200pi
If the base is 12 * 8, then volume is pi(4^2)(10) = 160pi

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?

got the same answer. Did this come from MGMAT?

I think I saw this question on ETS retired paper test.... can't be sure though but it does seem familiar to me.

Re: volume question [#permalink]
24 Oct 2007, 18:44

bmwhype2 wrote:

bmwhype2 wrote:

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

all right. i finally figured it out after some basic reasoning. a 12 inch ball cannot fit into a 12 inch box without some slack space.

therefore, we take the two largest dimensions as the base and then take the smaller of the base dimensions as a radius. thereby leaving the shortest dimension of the three as the height.

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