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Circles & Cylinders

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by adr » Tue Oct 20, 2009 5:22 am
A rectangular box has the dimensions 12 inches X 10 inches X 8 inches. What is the largest possible volume of a right cylinder that is placed inside the box?

Need help in solving this! Found this prob in the Manhattan Strategy Guide for Geometry.
-ADR
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by xcusemeplz2009 » Tue Oct 20, 2009 5:32 am
possible volumes(max) of the cylinder are 360pi,288pi,300pi,200pi,192pi and 160pi

therefore
largest possible volume is 360pi

we can get the value by using diff combinations of r and h from(12,10,8)


OA pls
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by adr » Tue Oct 20, 2009 6:40 am
Here's is the Book's Explanation:

"The Radius of the cylinder must be equal to half of the smaller 2 dimension that form the box's bottom. The height, then can be equal to the remaining dimension of the box"

Since V = pi.r^2.h

Case1: r = 5, h = 8
V = 25pi.8 = 200pi

Case2: r = 4, h = 10
V = 16pi.10 = 160pi

Case3: r = 4, h = 12
V = 16pi.12 = 192pi

Hence, Case 1 yields the larges volume ie 200pi "


But, if one considers the cases, we get 3 combinations ie:
12X8
12X10
10X8

Hence 12X10; R = 5 then V = 12.25pi = 300pi should be greatest volume? I guess the book is wrong in the case 1 in choosing its h and l.
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by pharmxanthan » Fri May 28, 2010 6:17 pm
What is a good way to select combinations of base and height in these types of questions?
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by liferocks » Fri May 28, 2010 6:27 pm
volume of the cylindr is pi*r^2*h where r is radius of base and h is height

so volume V is directly proportional to square of base radius and linearly proportional with height.
hence to ge max volume base radius has to be maximum and height to be second max

now th sides of the box are 12,10 and 8...clearly the max circle can be drawn in any of the sides will have radius 10/2..so base radius will be 10/2 and this is the side 12x10..so height has to be 8.hence volumn
=pi*5^2*8=200pi
Last edited by liferocks on Sat May 29, 2010 7:31 pm, edited 1 time in total.
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by gmatmachoman » Fri May 28, 2010 10:44 pm
liferocks wrote:volume of the cylindr is pi*r^2*h where r is radius of base and h is height

so volume V is directly proportional to square of base radius and linearly proportional with height.
hence to ge max volume base radius has to be maximum and height to be second max

now th sides of the box are 12,10 and 8...clearly the max circle can be drawn in any of the sides will have radius 10..so base radius will be 10 and this is the side 12x10..so height has to be 8.hence volumn
=pi*5^2*8=200pi
LR, in that case radius needs to be 12/2...rite?ia m confused!
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by gmatjedi » Sat May 29, 2010 2:46 am
for the 10x12 rectangular face, the height of the cylinder is 8, the radius is 5 (smaller of the 10x12)
for the 8x10 rectangular face, the height of the cylinder is 12, the radius is 4 (smaller of the 8x10)
for the 8x12 rectagular face, the height of the cylinder is 10, the radius is 4 (smaller of the 8x12)

it helps me to understand the above, by drawing the picture; then it's easier to know what is the height for a particular face

macho, in lr's calcs, to get 200 pi, he is using the first example above; in that case the circle's radius is limited by the smaller side of the rectangular face and the corresponding ht is 8.

i hope this helps
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by akhpad » Sat May 29, 2010 5:54 am
adr wrote:
But, if one considers the cases, we get 3 combinations ie:
12X8
12X10
10X8

Hence 12X10; R = 5 then V = 12.25pi = 300pi should be greatest volume? I guess the book is wrong in the case 1 in choosing its h and l.
What 12 here? it should 8.

200 pie is the max area
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by liferocks » Sat May 29, 2010 7:30 pm
gmatmachoman wrote:
liferocks wrote:volume of the cylindr is pi*r^2*h where r is radius of base and h is height

so volume V is directly proportional to square of base radius and linearly proportional with height.
hence to ge max volume base radius has to be maximum and height to be second max

now th sides of the box are 12,10 and 8...clearly the max circle can be drawn in any of the sides will have radius 10..so base radius will be 10 and this is the side 12x10..so height has to be 8.hence volumn
=pi*5^2*8=200pi
LR, in that case radius needs to be 12/2...rite?ia m confused!
the three sides will be 12x10,12x8 and 10x8..now in any rectangle the maximum circle can inscribe will have the diameter equal to smaller side.So in these three sides the largest inscribed circles will have radius 10/2,8/2 and 8/8 respectively..so max is 10/2=5
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