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A certain rectangular crate measures 8 feet by 10 feet by 12 feet. A c

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A certain rectangular crate measures 8 feet by 10 feet by 12 feet. A c [#permalink] New post 06 Nov 2009, 06:01
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A certain rectangular crate measures 8 feet by 10 feet by 12 feet. A cylindrical gas tank is to be made for shipment in the crate and will stand upright when the crate is placed on one of its six faces. What should the radius of the tank be if it is to be of the largest possible volume?

A. 4
B. 5
C. 6
D. 8
E. 10

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-certain-rectangular-crate-measures-8-feet-by-10-feet-by-100223.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 11 Sep 2014, 01:08, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: A certain rectangular crate measures 8 feet by 10 feet by 12 feet. A c [#permalink] New post 06 Nov 2009, 07:08
IndianGuardian wrote:
Came by this question in PR but unable to understand the solution:

A certain rectangular crate measures 8 feet by 10 feet by 12 feet. A cylindrical gas tank is to be made for shipment in the crate and will stand upright when the crate is placed on one of its six faces. What should the radius of the tank be if it is to be of the largest possible volume?

i) 4
ii) 5
iii) 6
iv) 8
v) 10

How do we get the solution as 5?

Thanks in advance.


We are not told the l, w, or h of the crate. Therefore the dimensions of the crate can be:

L x w x h = 8 x 10 x 12
l x w x h = 8 x 12 x 10
l x w x h = 10 x 12 x 8
l x w x h = 12 x 8 x 10
l x w x h = 10 x 8 x 12
l x w x h = 12 x 10 x 8

volume of cylinder = (pi)(r^2) * h

Let's take the 1st one: 8 x 10 x 12
If the cylinder is to be placed in the crate the largest the diameter can be is 8. It can't be 10 because one side is 10 and the other side is 8. The radius would be 4 and the volume would be: 192 pi

looking at the second set of numbers 8 x 12 x 10 largest diameter is 8 so no point of calculating because height will be 10 and will be smaller then 8 x 10 x 12

Do that for all of them and you get radius = 5 which is a 10 x 12 x 8 or 200 pi

You can quickly eliminate any dimension that has length or width as 8 because we already maximized volume with the 1st dimension. You would have quickly noticed that the next best option would have been a height of 8. Remember you don't have to work them all because 10 x 12 x 8 is the same as 12 x 10 x 8 because the diameter is the smaller of the length or width
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Re: A certain rectangular crate measures 8 feet by 10 feet by 12 feet. A c [#permalink] New post 06 Nov 2009, 07:16
IndianGuardian wrote:
Came by this question in PR but unable to understand the solution:

A certain rectangular crate measures 8 feet by 10 feet by 12 feet. A cylindrical gas tank is to be made for shipment in the crate and will stand upright when the crate is placed on one of its six faces. What should the radius of the tank be if it is to be of the largest possible volume?

i) 4
ii) 5
iii) 6
iv) 8
v) 10

How do we get the solution as 5?

Thanks in advance.


The cylinder's circular face should fit inside the crate's rectangular face.
Following possibilities whichever face we choose:
Crate-face dimensions............Crate Height.......Cylinder circle MAXIMUM radius possible in that face, area, VOL
8x10...................................12.....................8/pi x 16/pi x 16 x 12 = 192pi
8x12...................................10.....................8/pi x 16/pi x 16 x 10 = 160pi
10x12..................................8......................10/pi x 25/pi x 25 x 8 = 200pi

So the MAX is when we use the rectangular face 10x12 -> diameter 10 -> radius 5. B.
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largest volume [#permalink] New post 07 Oct 2011, 10:54
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A certain rectangular crate measures 8 feet by 10 feet by 12 feet. A cylindrical gas tank is to be made for shipment in the crate and will stand upright when the crate is placed on one of its six faces. What should the radius of the tank be if it is to be of the largest possible volume?

a. 4
b. 5
c. 6
d. 8
e. 10
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Re: largest volume [#permalink] New post 07 Oct 2011, 12:45
Hi,

This is my first post and response to any question in this forum.

Solution : To get the maximum base of the cylinder the crate has to be of dimension 10*12. Now the challenge is to identify what is lenght and what is width.

If 12 is the diameter of the cylinder then cylinder of equal proportion will not fit in one side of crate. So diameter has to be 10 so that cylinder will fit in the base of crate.

Radius will be 5 So answer is 5.

Thanks
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Re: largest volume [#permalink] New post 08 Oct 2011, 00:26
Vol= k*r^2*h

Dimensions are important to avoid confusion.
you can't have a 12*8 base as the circle won't fit in
so 10*12 is the best option
12 can't be an option for diameter.
so the diameter is 10
Hence radius is 5.
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Re: largest volume [#permalink] New post 09 Oct 2011, 20:53
blink005 wrote:
Vol= k*r^2*h

Dimensions are important to avoid confusion.
you can't have a 12*8 base as the circle won't fit in
so 10*12 is the best option
12 can't be an option for diameter.
so the diameter is 10
Hence radius is 5.


Thre is a confusion here ..
i think in a 12*8 base circle will fit in if the radius is 4.
We cannot have 12 as diameter because it will not fit as other sides are smaller than this.
So it is between 12*10 and 12*8 with 10 and 8 diameters respectively.

but we need maximum volume so it should be 12*10 with 10 as diameter
ans r = 5
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Re: largest volume [#permalink] New post 11 Oct 2011, 23:01
Hey '

Can you explain why diameter of 12 is not workable ?
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Re: largest volume [#permalink] New post 12 Oct 2011, 01:59
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Priyanka2011 wrote:
Hey '

Can you explain why diameter of 12 is not workable ?

As per the question the cylinder should fit in the crate.

If The dimension of the base is l*b then the diameter of the circle should be equal or less than the smaller side.Otherwise the it will not fit in. it will look lik below image.

Here we have three options 12X10 or 12X8 or 10 X 8 of all these 12 X 10 will give the largest volume.
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2.JPG [ 8.93 KiB | Viewed 1540 times ]


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Re: largest volume [#permalink] New post 23 Oct 2011, 01:53
the area or volume are maximum when the dimensions are as close to each other as possible as is the case in square or cube. so, 12x10 would be the dimensions of the cylinder. a circle only fits perfectly inside of a square (inscribed) not a rectangle, so d=12 will not fit 12x10. therefore, d has to be 10, to fit inside 12x10, and h=12 to make the \(v=300\pi\).
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Re: A certain rectangular crate measures 8 feet by 10 feet by 12 feet. A c [#permalink] New post 10 Sep 2014, 22:04
gmattokyo wrote:
IndianGuardian wrote:
Came by this question in PR but unable to understand the solution:

A certain rectangular crate measures 8 feet by 10 feet by 12 feet. A cylindrical gas tank is to be made for shipment in the crate and will stand upright when the crate is placed on one of its six faces. What should the radius of the tank be if it is to be of the largest possible volume?

i) 4
ii) 5
iii) 6
iv) 8
v) 10

How do we get the solution as 5?

Thanks in advance.


The cylinder's circular face should fit inside the crate's rectangular face.
Following possibilities whichever face we choose:
Crate-face dimensions............Crate Height.......Cylinder circle MAXIMUM radius possible in that face, area, VOL
8x10...................................12.....................8/pi x 16/pi x 16 x 12 = 192pi
8x12...................................10.....................8/pi x 16/pi x 16 x 10 = 160pi
10x12..................................8......................10/pi x 25/pi x 25 x 8 = 200pi

So the MAX is when we use the rectangular face 10x12 -> diameter 10 -> radius 5. B.


@ Bunuel

Hi Bunuel, i have a doubt in this above question. Wht cant the radius by 6 (12/2). Why cant we take 12 as length of the face. Please clarify.
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Re: A certain rectangular crate measures 8 feet by 10 feet by 12 feet. A c [#permalink] New post 10 Sep 2014, 22:56
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IndianGuardian wrote:
Came by this question in PR but unable to understand the solution:

A certain rectangular crate measures 8 feet by 10 feet by 12 feet. A cylindrical gas tank is to be made for shipment in the crate and will stand upright when the crate is placed on one of its six faces. What should the radius of the tank be if it is to be of the largest possible volume?

i) 4
ii) 5
iii) 6
iv) 8
v) 10

How do we get the solution as 5?

Thanks in advance.


The dimensions of the crate are 8*10*12. The rectangular base of the crate can be 8*10 or 10*12 or 8*12 and the height will be the leftover dimension.
To maximize volume of the cylinder, we must maximize radius and height. The diameter of the cylinder will be the measure of the shorter side of the base i.e. if the base of the crate measures 8*10, the diameter will be 8. So one thing is clear - the two sides of the crate should be as close as possible to each other in measure because some space gets wasted. Hence having the base as 8*12 and height as 10 doesn't make sense. It is much better to have base as 8*10 and height as 12 since diameter will be 8 in both cases but height will be 12 in the second case.

Now, radius gets squared so larger the radius, more impact it will have on volume as compared with height. But radius is half of diameter so some impact is lost. Let's review both leftover cases:

Base 8*10, height 12
Volume of cylinder \(= \pi*4^2*12 = 192*\pi\)
.
Base 10*12, height 8
Volume of cylinder \(= \pi*5^2*8 = 200*\pi\)

The volume will be maximum when base is 10*12 (so radius is 5).

Answer (B)
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Re: A certain rectangular crate measures 8 feet by 10 feet by 12 feet. A c [#permalink] New post 11 Sep 2014, 01:09
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IndianGuardian wrote:
A certain rectangular crate measures 8 feet by 10 feet by 12 feet. A cylindrical gas tank is to be made for shipment in the crate and will stand upright when the crate is placed on one of its six faces. What should the radius of the tank be if it is to be of the largest possible volume?

A. 4
B. 5
C. 6
D. 8
E. 10



Volume of the cylinder equals to \(area=\pi{r^2}h\). First of all note that answer choices C, D, and E don't make sense. For example cylinder of a radius 6 (option C) just won't fit on any face, as max face has a dimensions 12*10 so cylinder with max radius of 5 can be placed on it.

Max volume will be when the base of a cylinder is placed on the face with dimension 12*10 thus the radius will be 5 --> \(v=\pi{5^2}*8=200\pi\);

Other options:
If we place cylinder on the face with dimension 12*8 then radius will be 4 and \(v=\pi{4^2}*10=160\pi\);
If we place cylinder on the face with dimension 10*8 then radius will be 4 and \(v=\pi{4^2}*12=192\pi\),.

Answer: B.

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Re: largest volume [#permalink] New post 27 Apr 2015, 20:50
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Re: largest volume   [#permalink] 27 Apr 2015, 20:50
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