The inside dimensions of a rectangular wooden box
are 6 inches by 8 inches by 10 inches. A cylindrical
canister is to be placed inside the box so that it stands
upright when the closed box rests on one of its six
faces. Of all such canisters that could be used, what is
the radius, in inches, of the one that has the
maximum volume?
A-3
B-4
C-5
D-6
E-8
In this question Area could be max when radius = 5 and height = 8
A= 3.14 *25*8 = 100 * 3.14
Is it ? Then answer should be C but OA is B.
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MartyMurray
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If the radius were 5, the cylinder would not fit inside the box.
The biggest face of the box is 8 x 10.
A circle with a radius of 5 has a diameter of 10, or is 10 wide, in all directions. So, a cylinder with a radius 5 would therefore not fit in a box unless at least one face of that box were to have dimensions of at least 10 x 10.
Given that you are constrained by either width 8, on the 8 x 10 end, or width 6, on the 6 x 8 or the 6 x 10 end, the greatest two diameters possible are 8 and 6, and the greatest two radii possible are 8/2 = 4 and 6/2 = 3.
Formula for the volume of a cylinder: π x r² x Height
Radius 3 on the 6 x 10 end generates π x 3² x 8 = 72π (You don't really need to calculate this one as the next one will have the same radius and a greater height, and so the next will will obviously have a greater volume.)
Radius 3 on the 6 x 8 end generates π x 3² x 10 = 90π
Radius 4 on the 8 x 10 end generates π x 4² x 6 = 96π
So the correct answer is B.
The biggest face of the box is 8 x 10.
A circle with a radius of 5 has a diameter of 10, or is 10 wide, in all directions. So, a cylinder with a radius 5 would therefore not fit in a box unless at least one face of that box were to have dimensions of at least 10 x 10.
Given that you are constrained by either width 8, on the 8 x 10 end, or width 6, on the 6 x 8 or the 6 x 10 end, the greatest two diameters possible are 8 and 6, and the greatest two radii possible are 8/2 = 4 and 6/2 = 3.
Formula for the volume of a cylinder: π x r² x Height
Radius 3 on the 6 x 10 end generates π x 3² x 8 = 72π (You don't really need to calculate this one as the next one will have the same radius and a greater height, and so the next will will obviously have a greater volume.)
Radius 3 on the 6 x 8 end generates π x 3² x 10 = 90π
Radius 4 on the 8 x 10 end generates π x 4² x 6 = 96π
So the correct answer is B.
Last edited by MartyMurray on Wed Nov 25, 2015 7:52 am, edited 2 times in total.
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Brent@GMATPrepNow
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Volume of cylinder = pi(radius^2)(height)vishalwin wrote:The inside dimensions of a rectangular wooden box
are 6 inches by 8 inches by 10 inches. A cylindrical
canister is to be placed inside the box so that it stands
upright when the closed box rests on one of its six
faces. Of all such canisters that could be used, what is
the radius, in inches, of the one that has the
maximum volume?
A-3
B-4
C-5
D-6
E-8
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 6x8 side, on the 6x10 side, or on the 8x10 side
If you place the base on the 6x8 side, then the cylinder will have height 10, and the maximum radius of the cylinder will be 3 (i.e., diameter of 6).
So, the volume of this cylinder will be (pi)(3^2)(10), which equals 90(pi)
If you place the base on the 6X10 side, then the cylinder will have height 8, and the maximum radius of the cylinder will be 3 (i.e., diameter of 6).
So, the volume of this cylinder will be (pi)(3^2)(8), which equals 72(pi)
If you place the base on the 8x10 side, then the cylinder will have height 6, and the maximum radius of the cylinder will be 4 (i.e., diameter of 8).
So, the volume of this cylinder will be (pi)(4^2)(6), which equals 96(pi)
So, the greatest possible volume is 96(pi) and this occurs when the radius is 4
Cheers,
Brent
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Hi vishalwin,
You have to be careful with how you set-up this question (drawing pictures would likely help though). Since none of the sides of the box is a square, the diameter of the cylinder is limited by the SHORTER dimension of the side that is face down.
For example, if the 6x8 side was face down, then the cylinder would have a maximum diameter of 6. That limitation holds true with the other two options as well (6x10 face down ---> max diameter = 6; 8x10 face down --> max diameter = 8). Keeping that in mind, what answer would you get if you solved as normal?
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You have to be careful with how you set-up this question (drawing pictures would likely help though). Since none of the sides of the box is a square, the diameter of the cylinder is limited by the SHORTER dimension of the side that is face down.
For example, if the 6x8 side was face down, then the cylinder would have a maximum diameter of 6. That limitation holds true with the other two options as well (6x10 face down ---> max diameter = 6; 8x10 face down --> max diameter = 8). Keeping that in mind, what answer would you get if you solved as normal?
GMAT assassins aren't born, they're made,
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Amrabdelnaby
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Vishalwin,
the cylinder could be placed on one of three sides: 6x8 , 6x10 , 8x10
however you need to notice that the diameter of the cylinder will be constrained by the smaller side
for example if it is placed on the side 6x8, the diameter can't be 8, because the other dimension of the box is 6, so the cylinder won't fit in; hence, we are constrained by the smaller side in every case of the three cases.
so lets consider case 1: 6x8
maximum diameter here will be 6 so radius is 3, and height is the other dimension of the box which is 10
so the volume will be 9pix10 = 90 pi
case 2: 6x10
we are constrained by the smaller dimension, 6, which will be the maximum diameter for the cylinder.
radius will still be 3, so volume will be 9pi x 8 = 72pi --> here 8 is the third dimension of the box
case 3: 8x10
the maximum diameter can be 8, so radius will be 4
volume equal: 16pi x 6 = 92pi
i hope you got it
cheers
the cylinder could be placed on one of three sides: 6x8 , 6x10 , 8x10
however you need to notice that the diameter of the cylinder will be constrained by the smaller side
for example if it is placed on the side 6x8, the diameter can't be 8, because the other dimension of the box is 6, so the cylinder won't fit in; hence, we are constrained by the smaller side in every case of the three cases.
so lets consider case 1: 6x8
maximum diameter here will be 6 so radius is 3, and height is the other dimension of the box which is 10
so the volume will be 9pix10 = 90 pi
case 2: 6x10
we are constrained by the smaller dimension, 6, which will be the maximum diameter for the cylinder.
radius will still be 3, so volume will be 9pi x 8 = 72pi --> here 8 is the third dimension of the box
case 3: 8x10
the maximum diameter can be 8, so radius will be 4
volume equal: 16pi x 6 = 92pi
i hope you got it
cheers
vishalwin wrote:The inside dimensions of a rectangular wooden box
are 6 inches by 8 inches by 10 inches. A cylindrical
canister is to be placed inside the box so that it stands
upright when the closed box rests on one of its six
faces. Of all such canisters that could be used, what is
the radius, in inches, of the one that has the
maximum volume?
A-3
B-4
C-5
D-6
E-8
In this question Area could be max when radius = 5 and height = 8
A= 3.14 *25*8 = 100 * 3.14
Is it ? Then answer should be C but OA is B.
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The idea is right, but try drawing a circle with diameter 10 inside an 8x10 rectangle and you'll see what goes wrong. I run into this issue myself sometimes, and drawing a figure to scale usually clears things up.
