circles and cylinders

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circles and cylinders

by abhinav khanna » Thu Feb 02, 2012 8:02 am
a rectangular box has the dimensions 12x10x8 inches. What is the largest possible volume of a right cylinder that can be placed in the box?

Kindly explain what does right cylinder mean?

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by Brent@GMATPrepNow » Thu Feb 02, 2012 8:12 am
abhinav khanna wrote:a rectangular box has the dimensions 12x10x8 inches. What is the largest possible volume of a right cylinder that can be placed in the box?

Kindly explain what does right cylinder mean?

Regards
It's a cylinder that you're used to. When we set it on the ground, it stands at a 90-degree angle to the ground.
Most people would just call this a cylinder, but the GMAT wants to be 100% mathematically correct, so that say "right cylinder" to rule out cylinders that stand at an angle that is not 90 degrees (kind of like the Leaning Tower of Pisa)

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by abhinav khanna » Thu Feb 02, 2012 8:34 am
Thanks Brent for the quick response, could you guide me to the max volume that a right cylinder can have in the said box.

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by Brent@GMATPrepNow » Thu Feb 02, 2012 9:24 am
abhinav khanna wrote:Thanks Brent for the quick response, could you guide me to the max volume that a right cylinder can have in the said box.

Regards
Sure thing.

This one is tough to visualize though.

There are 3 different ways to position the cylinder (with the base on a different side each time).
You can place the base on the 12x10 side, on the 10x8 side, and on the 12x8 side

If you place the base on the 12x10 side, then the cylinder will have height 8, and the maximum radius of the cylinder will be 5.
So, the volume of this cylinder will be (pi)(5^2)(8), which equals 200(pi)

If you place the base on the 10x8 side, then the cylinder will have height 12, and the maximum radius of the cylinder will be 4.
So, the volume of this cylinder will be (pi)(4^2)(12), which equals 192(pi)

If you place the base on the 12x8 side, then the cylinder will have height 10, and the maximum radius of the cylinder will be 4.
So, the volume of this cylinder will be (pi)(4^2)(10), which equals 160(pi)

So, the greatest possible volume is 200pi

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by abhinav khanna » Thu Feb 02, 2012 11:14 am
Thanks Brent for the guidance! Actually i was going with this logic that Volume = Pi r^2 h, hence volume will be max when r and h have greatest values and i took r=6 and h=10 for that matter.

Now the doubt is cleared

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by ranvijay87 » Fri Feb 03, 2012 11:44 am
could you please clarify the doubt mentioned in the above post as i
too seem to be getting 360pi as the max value.
height = 10 and radius = 6.

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by Brent@GMATPrepNow » Fri Feb 03, 2012 12:02 pm
ranvijay87 wrote:could you please clarify the doubt mentioned in the above post as i
too seem to be getting 360pi as the max value.
height = 10 and radius = 6.
If a cylinder with radius 6 and height 10 is placed in the box, then the circular base of the cylinder must be on the side with dimensions 12x8 (in order for the height to be 10)

Now if the circular base is placed on the side with dimensions 12x8, what is the maximum length of the radius?

Well, a radius of 4 would mean a diameter of 8, which means the circular base would already be touching two sides of the box (since one side has length 8)

So, the maximum length of the radius is 4. Any circle with a radius greater that 4 would not fit on the 12x8 side.

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by ronnie1985 » Sat Feb 04, 2012 3:07 am
Right cylinder is the cylinder whose axis is perpendicular to its base.
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by somsubhra86 » Sun Feb 05, 2012 5:57 am
@Brent:-- I have doubt that is why are you taking slant height =10 not 12. If we take 12 as slant we will gt higher volume.

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by Brent@GMATPrepNow » Sun Feb 05, 2012 7:32 am
somsubhra86 wrote:@Brent:-- I have doubt that is why are you taking slant height =10 not 12. If we take 12 as slant we will gt higher volume.
Don't forget that we're dealing with a right cylinder here.
So, if we want it to have a height of 12, we must its cicular base on the 10x8 side.
In this case, the maximum radius of the cylinder will be 4.
So, the volume of this cylinder will be (pi)(4^2)(12), which equals 192(pi)

This is not the greatest volume (see above)

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