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A rectangular box has dimensions 12*10*8 inches. What is the
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02 Mar 2012, 03:19
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Ashamock wrote:
Why cant we have radius =6 and height = 10 making the volume 360 Pi ?
Dimensions of the box are 12*10*8 inches if radius of a cylinder is 6 then its diameter is 12 and it won't fit on any face of a box. For example it can not fit on 12*10 face of the box since diameter=12>10=side.
Complete solution: A rectangular box has dimensions 12*10*8 inches. What is the largest possible value of right cylcinder that can be placed inside the box?
\(volume_{cylinder}=\pi{r^2}h\)
If the cylinder is placed on 8*10 face then it's maximum radius is 8/2=4 and \(volume==\pi*{4^2}*12=192\pi\); If the cylinder is placed on 8*12 face then it's maximum radius is 8/2=4 and \(volume==\pi*{4^2}*10=160\pi\); If the cylinder is placed on 10*12 face then it's maximum radius is 10/2=5 and \(volume==\pi*{5^2}*8=200\pi\);
Re: A rectangular box has dimensions 12*10*8 inches. What is the
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Updated on: 18 Jul 2013, 03:19
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Since the radius of the cylinder is squared to obtain its volume, the radius should be the maximum possible radius in order for the cylinder to achieve its maximum volume. Point to keep in mind while selecting the maximum possible radius: The diameter should be <= two of the largest sides of the rectangle. If one selects the largest side of the rectangle as the diameter, then the cylinder won't fit into the other side of the rectangle. Refer Figure.
Attachment:
cylinder in rectangle.JPG [ 10.15 KiB | Viewed 29111 times ]
As a rule of thumb, in such problems, select the second largest side as the diameter (note that it is the diameter and one has to calculate the radius by dividing by 2 before calculating the volume). And the left alone smallest side will be the height of the cylinder (as you need the two largest sides to enclose the bottom of the cylinder the only choice left out for height is the smallest side).
Originally posted by Jaisri on 18 Jul 2013, 03:15.
Last edited by Jaisri on 18 Jul 2013, 03:19, edited 1 time in total.
(If h=12, r can't be more than 4. So pi*r^2*h=12*16pi=192pi
Apparently h=10, r=4 can't be more
Since vol. involves r^2, looking for a greater r would help. However r can't be 6 because the max possible length of the other side could be only 10 which could not fit a circular base)
Remember: the base of the cylinder is a circle
r is constant around the circle, so the base can completely fit in a square, not a rectangle. Therefore, you will not be able to use the entire space in the base of the rectangular box.
the volume of a cylinder is pi * r^2 * h
select largest possible r such that the dimensions of the rectangular base allow maximum value or r
you would want to maximize r more than h because r is squared here
so, select bases 10 and 12 allowing a maximum value of r = 5
volume of cylinder: 25 pi * 8 = 200 pi
to check, think if 8 is one of the dimensions of the base, r = 4
volume of cyclinder: 16 pi * (either 10 or 12 ) = 160 pi or 192 pi
not maximum volume
Why cant we have radius =6 and height = 10 making the volume 360 Pi ?
Dimensions of the box are 12*10*8 inches if radius of a cylinder is 6 then its diameter is 12 and it won't fit on any face of a box. For example it can not fit on 12*10 face of the box since diameter=12>10=side.
Complete solution: A rectangular box has dimensions 12*10*8 inches. What is the largest possible value of right cylcinder that can be placed inside the box?
\(volume_{cylinder}=\pi{r^2}h\)
If the cylinder is placed on 8*10 face then it's maximum radius is 8/2=4 and \(volume==\pi*{4^2}*12=196\pi\); If the cylinder is placed on 8*12 face then it's maximum radius is 8/2=4 and \(volume==\pi*{4^2}*10=160\pi\); If the cylinder is placed on 10*12 face then it's maximum radius is 10/2=5 and \(volume==\pi*{5^2}*8=200\pi\);
Why cant we have radius =6 and height = 10 making the volume 360 Pi ?
Dimensions of the box are 12*10*8 inches if radius of a cylinder is 6 then its diameter is 12 and it won't fit on any face of a box. For example it can not fit on 12*10 face of the box since diameter=12>10=side.
Complete solution: A rectangular box has dimensions 12*10*8 inches. What is the largest possible value of right cylcinder that can be placed inside the box?
\(volume_{cylinder}=\pi{r^2}h\)
If the cylinder is placed on 8*10 face then it's maximum radius is 8/2=4 and \(volume==\pi*{4^2}*12=196\pi\); If the cylinder is placed on 8*12 face then it's maximum radius is 8/2=4 and \(volume==\pi*{4^2}*10=160\pi\); If the cylinder is placed on 10*12 face then it's maximum radius is 10/2=5 and \(volume==\pi*{5^2}*8=200\pi\);
Re: A rectangular box has dimensions 12*10*8 inches. What is the
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08 Oct 2013, 07:15
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willget800 wrote:
A rectangular box has dimensions 12*10*8 inches. What is the largest possible value of right cylinder that can be placed inside the box?
A. 180 pie B. 200 Pie C. 300 Pie D. 320 Pie E. 450 Pie
Always remember, the largest possible value will have a diameter that will need to feet on 2 of the sides. Therefore, if it fits in 10 it will also fit in 12. So choose a diameter of 10 = 2r, r=5 and then use the other side 8 for the height giving a total volume of 200pi
Re: A rectangular box has dimensions 12*10*8 inches. What is the
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27 Dec 2013, 16:52
willget800 wrote:
A rectangular box has dimensions 12*10*8 inches. What is the largest possible value of right cylinder that can be placed inside the box?
A. 180 pie B. 200 Pie C. 300 Pie D. 320 Pie E. 450 Pie
So first we need to place the cylinder so that its diameter is >= to at least two of the dimensions. That is 10 and 12, so r = 5
Next, height has to be 8 then. I suggest you draw the cuboid and see for yourself that we need a side where the circumference stands to be 10*12 otherwise diameter won't fit
Then I guess the question asks for the largest volume possible
Re: A rectangular box has dimensions 12*10*8 inches. What is the
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14 Oct 2015, 03:23
Bunuel wrote:
Ashamock wrote:
Why cant we have radius =6 and height = 10 making the volume 360 Pi ?
Dimensions of the box are 12*10*8 inches if radius of a cylinder is 6 then its diameter is 12 and it won't fit on any face of a box. For example it can not fit on 12*10 face of the box since diameter=12>10=side.
Complete solution: A rectangular box has dimensions 12*10*8 inches. What is the largest possible value of right cylcinder that can be placed inside the box?
\(volume_{cylinder}=\pi{r^2}h\)
If the cylinder is placed on 8*10 face then it's maximum radius is 8/2=4 and \(volume==\pi*{4^2}*12=196\pi\); If the cylinder is placed on 8*12 face then it's maximum radius is 8/2=4 and \(volume==\pi*{4^2}*10=160\pi\); If the cylinder is placed on 10*12 face then it's maximum radius is 10/2=5 and \(volume==\pi*{5^2}*8=200\pi\);
Re: A rectangular box has dimensions 12*10*8 inches. What is the
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04 Aug 2016, 04:04
a doubt on basics : it is a very silly question. But why are we considering volume and not surface area ? I got confused and made a blunder in the problem
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