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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
(A) 3
(B) 4
(C) 5
(D) 6
(E) 8
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BANON
\(volume_{cylinder}=\pi{r^2}h\)
If the cylinder is placed on 6*8 face then it's maximum radius is 6/2=3 and \(volume==\pi*{3^2}*10=90\pi\);
If the cylinder is placed on 6*10 face then it's maximum radius is 6/2=3 and \(volume==\pi*{3^2}*8=72\pi\);
If the cylinder is placed on 8*10 face then it's maximum radius is 8/2=4 and \(volume==\pi*{4^2}*6=96\pi\);
So, the maximum volume is for \(r=4\).
Answer: B.
Hope it's clear.
P.S. Notice that radii of 5, 6, and 8 (options C, D and E) are not possible.
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The three different sizes of faces of the rectangular box are:
6 x 8
8 x 10
6 x 10
Now in simple terms the question asks us to find the maximum radius of the cylinder which can be placed inside the box (reference to volume is an unwanted piece of information to solve this question)
The cylinder with the longest radius can logically sit only on the face of the box with the largest area which is 8 x 10.
This face of the box can accommodate a cylinder with diameter equal to the rectangular face's shortest length i.e 8. hence the maximum possible radius for the cylinder will be 4.
Hope the above is clear.
6 x 8
8 x 10
6 x 10
Now in simple terms the question asks us to find the maximum radius of the cylinder which can be placed inside the box (reference to volume is an unwanted piece of information to solve this question)
The cylinder with the longest radius can logically sit only on the face of the box with the largest area which is 8 x 10.
This face of the box can accommodate a cylinder with diameter equal to the rectangular face's shortest length i.e 8. hence the maximum possible radius for the cylinder will be 4.
Hope the above is clear.
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Found this useful tip to be remembered as a thumb rule in another post .
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).
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).
General Discussion
Posts: 2
So as a general rule, can we say that the longest side (10 in this case) can never be the Diameter of the circular base? I am having difficulty visualizing the reason of 10 not being a viable option for diameter.
planesmypassion
Ask yourself: on which face of the cube can you place a cylinder with the diameter of 10.
Even if you consider the largest face (8*10), it's one dimension (8) will still be less than the diameter, so you cannot place the cylinder on it.
Hope it's clear.
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Posts: 2
Bunuel
Drew a few figures and I think I understand now. When calculating , we can never take the larger side as diameter as the circular base would extend outside of the rectangular face.
And in this case , only 3 cases possible where dia<height. Thanks bro!!
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Are there any similar questions to this for practice purposes?
BANON
pinkcupcake
Similar questions to practice:
the-inside-of-a-rectangular-carton-is-48-centimeters-long-138301.html
a-rectangular-box-has-dimensions-12-10-8-inches-what-is-the-28790.html
a-closed-cylindrical-tank-contains-36pi-cubic-feet-of-water-134500.html
what-is-the-number-of-cans-that-can-be-packed-in-a-certain-143776.html
a-cylindrical-tank-of-radius-r-and-height-h-must-be-redesign-122366.html
a-rectangular-box-has-dimensions-12-10-8-inches-what-is-the-28790.html
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Bunuel
what would be the problem if volume= pi* 4^2* 8 ??
anik19890
hi anik19890, volume= \(pi* 4^2* 8\) is not possible because if height = 8, we are only left with dimensions 10 & 6.
Max of a cylinder's diameter from a base with dimensions 10 x 6 is 6, hence radius is \(6/2=3\).
If the height used is 8, then max volume of the cylinder is \(pi*3^2*8\) = \(72pi\)
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BANON
For the base of the crate we have three values to choose from and we need to select any two ==> 3C2=\(\frac{3!}{2!*1!}\) = 3 ways in which it can be done.
SO we will have three unique scenarios for the dimension of the base; the third value will become the height.
1st Case) Take base of 6 by 8; Radius 6/2=3 and the height 10 ; volume = 9*10 =90
2nd Case) Take base of 6 by 10; Radius 6/2=3 and the height 8; volume = 9*8= 72
3rd Case) Take base of 8 by 10, Radius 8/2=4 and the height 6; volume 16*6=96
The volume is greatest when radius =4
Hence B is the answer
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BANON
In order for the canister to stand upright in the box, the diameter of the canister must fit within the base of the box. Let’s test various scenarios to determine which will provide the largest volume of the canister. Remember, the volume of a cylinder = πr^2h. Keep in mind that the height of the cylindrical canister is the same as the height of the box.
Scenario 1:
The base of the box is 6 by 8 and the height is 10. Thus, the diameter of the cylinder = 6, which means the radius = 3.
V = π(3)^2 x 10 = 90π
Scenario 2:
The base of the box is 6 by 10 and the height is 8. Thus, the diameter of the cylinder = 6, which means the radius = 3.
V = π(3)^2 x 8 = 72π
Scenario 3:
The base of the box is 8 by 10 and the height is 6. Thus, the diameter of the cylinder = 8, which means the radius = 4.
V = π(4)^2 x 6 = 96π
Thus, the radius of the cylinder that provides the largest volume is 4.
Answer: B
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BANON
Volume of cylinder = pi(radius²)(height)
There are 3 different ways to position the cylinder (with the base on a different side each time).
You can place the flat BASE of the cylinder 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)(3²)(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)(3²)(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)(4²)(6), which equals 96(pi)
So, the greatest possible volume is 96(pi) and this occurs when the radius is 4
Answer: B
Cheers,
Brent
volume of Cuboid = 480
To get placed inside the cuboid
Volume of cuboid= volume of cylinder
pi r^2 h = 480
Concept of solving the unknown
To stand upright, the cylinder canister should have a height equal to the cuboid largest side (i.e= 10)
solving for above eq.
r~4
(All dimensions are in inches and volume inches^3)
To get placed inside the cuboid
Volume of cuboid= volume of cylinder
pi r^2 h = 480
Concept of solving the unknown
To stand upright, the cylinder canister should have a height equal to the cuboid largest side (i.e= 10)
solving for above eq.
r~4
(All dimensions are in inches and volume inches^3)
I understand that the diameter h=10,(r=5) cylinder cannot be put in the rectangular box.
6 can be put(r=3)
But how come diameter 8 (r=4) can be fit into the cylinder.because that is not going to fit into the mouth of rectangular box, in my visualization i can fit only r=3 ( d=6)....
just think about a circle in square.. the circle in square it touches all sides thats why diameter is equal to all 4 sides.
But when it comes for rectangular side the diameter is not equal to all 4 sides. The mouth of the box is narrow in rectangular because of lenght and width.
So how come you can put 2 diameter(8,6) and it touches the all sides in the rectangular box just like square mouth???
Need help not getting this thing in my head
Posted from my mobile device
6 can be put(r=3)
But how come diameter 8 (r=4) can be fit into the cylinder.because that is not going to fit into the mouth of rectangular box, in my visualization i can fit only r=3 ( d=6)....
just think about a circle in square.. the circle in square it touches all sides thats why diameter is equal to all 4 sides.
But when it comes for rectangular side the diameter is not equal to all 4 sides. The mouth of the box is narrow in rectangular because of lenght and width.
So how come you can put 2 diameter(8,6) and it touches the all sides in the rectangular box just like square mouth???
Need help not getting this thing in my head
Posted from my mobile device
Bunuel
I understand that the diameter h=10,(r=5) cylinder cannot be put in the rectangular box.
6 can be put(r=3)
But how come diameter 8 (r=4) can be fit into the cylinder.because that is not going to fit into the mouth of rectangular box, in my visualization i can fit only r=3 ( d=6)....
just think about a circle in square.. the circle in square it touches all sides thats why diameter is equal to all 4 sides.
But when it comes for rectangular side the diameter is not equal to all 4 sides. The mouth of the box is narrow in rectangular because of lenght and width.
So how come you can put 2 diameter(8,6) and it touches the all sides in the rectangular box just like square mouth???
Need help not getting this thing in my head
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GMATPrepNow
Hi Brent,
I am not able to visualise this. how can we take each face say 6*8 and decide which will be the diameter
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Asad
Hi Asad
Kindly refer this statement from question.
" 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".
Now we can't have a cylinder completely rested inside box with r>=5.
The best scenario could be if box rested on face 10*8.
Here cylinder of radius r>=5 can't fit. hence only option are r=3 or r=4.
Hope it helps!
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