The outer dimensions of a closed rectangular cardboard box are 8 centi

Answer: Option C

Please check the video for the step-by-step solution.

GMATinsight's Solution

The inner dimensions of the rectangular box are 7 x 9 x 11 because we have to subtract the side thickness twice from each outer dimension.

Case 1: The dimensions of the base of the box are 7 x 9.

In this case, the maximum radius for the base of the cylinder is 3.5, and consequently the volume of the cylinder is

V = (3.5)^2(pi)(11) = 134.75(pi)

Case 2: The dimensions of the base of the box are 7 x 11.

In this case, the maximum radius for the base of the cylinder is again 3.5.

V = (3.5)^2(pi)(9) < 134.75(pi)

Case 3: The dimensions of the base of the box are 9 x 11.

In this case, the maximum radius for the base of the cylinder is 4.5.

V = (4.5)^2(pi)(7) = (20.25)(pi)(7) > 134.75(pi)

Answer: C

Why can't 11 be the base. Then the radius will be 5.5

If one dimension of the inner rectangular base of the box is 11, then the other dimension must be either 7 or 9. If the radius of the base of the cylinder is 5.5, then the cylinder cannot be placed inside the box to stand on that base because the rectangular box would be too narrow to contain the cylinder.

Dimensions available inside of the rectangular cardboard box are (8 - \(\frac{1}{2}\) - \(\frac{1}{2}\)), (10 - \(\frac{1}{2}\) - \(\frac{1}{2}\)) and (12 - \(\frac{1}{2}\) - \(\frac{1}{2}\)) i.e. 7, 9 and 11

Volume of right circular cylinder = \(π * r^2 * h\). Here, volume is a function of 'r' and 'h'.

Here if the effect of an increase in r is more than increase in h if increase is same for both. So, for Volume to be maximum the guiding parameter is radius 'r' beause degree of 'r' is more than degree of 'h'.

Possible dimension of 'r' an be chosen from three rectangular surfaces with dimensions 7,9; 9,11 and 7,11, lower dimension deciding the final dimension.
So, possible values of 'r' are \(\frac{7}{2}\), \(\frac{9}{2}\) and \(\frac{7}{2}\) and heights would be 11, 7 and 9 respectively.

But verifying it since height would differ in the three cases.
Volume for if h is 11 = \(π * (\frac{7}{2})^2 * 11 = \frac{439}{4}π\)

Volume for if h is 7 = \(π * (\frac{9}{2})^2 * 7 = \frac{567}{4}π\)

Volume for if h is 9 = \(π * (\frac{7}{2})^2 * 9 = \frac{441}{4}π\)

Hence required radius = \(\frac{9}{2}\) = 4.5

Answer C.
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Hey lnm87

Your answer would be wrong, if the outer dimensions of a closed rectangular cardboard box were 8 centimeters by 10 centimeters by 13 centimeters. (just a slight change)

Volume depends on \(d^2*h\)

2 cases possible-

1. \(d=9; (d^2*h)_{max}= 9^2*7 = 81*7\)

2.\( d=7; (d^2*h)_{max}= 7^2*12 = 84*7\)

Hence answer would be 3.5.

Let me make it very simple, it asked for radius of max. cylinder the box can contain.
Now the max volume of cylinder is close to the total volume of box.

So equate 10*8*12>pi*r^2*h
h=12

10*8>π*r^2
r^2<24
Approx . 4.5

Posted from my mobile device

We see that each of the inner dimensions of the box is 2 x ½ = 1 cm less than its respective outer dimension. Therefore, the inner dimensions of the box are 7 cm by 9 cm by 11 cm. Since the canister has to be able to stand upright when the box rests on one of its sides, the maximum value if the canister’s radius is 4.5 cm (notice that would make the diameter = 9 cm, equaling the smaller of the dimensions of the base of the box if the base is 9 cm by 11 cm and height is 7 cm).

Answer: C

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nick1816

Can we do it this way ..

If I visualize a rectangular box with usable sides of 7,9,11 as thickness is 1/2 cm on each side.

let the Length be 11 and depth be 9. The max diameter can be 9, as one side it is limiting the diameter, hence 4.5 radius

nick1816 Bunuel ScottTargetTestPrep

I usually try to avoid messy calculations and use intuition to get to the answer. And usually intuition says the term which is raised to a power (greater than 1) would be the higher value.

But if I have time to work out the numbers with approx values --> 3.5 to 4 and 4.5 to 5 then we get the following result

Case I: \(\pi.r^2.h = \pi.4^2.11 = 176\pi\)
Case II: \(\pi.r^2.h = \pi.4^2.9 = 144\pi\)
Case III: \(\pi.r^2.h = \pi.5^2.7 = 175\pi\)

This is erroneously showing that Case I is greater volume than Case III.

My question is when do we NOT approximate?

What i recommend is that if the options are not very close, only then go for approximation. In these kinda questions, in which we have no clue about how close the volume we'll get in 3 cases, look for other ways to save time.

For example- \(V= \pi.r^2.h\) or \(\pi.\frac{d^2}{4}.h\)
Volume is directly proportional to product of square of diameter and height.

We only need to calculate for 2 cases in this question

1. d=9 and h=7 ; d^2*h = 9^2*7 = 81*7

2. d=7 and h=11 ; d^2*h = 7^2*11 = 77*7

If the OUTER dimensions are 8 cm x 10 cm x 12 cm, and each side is 0.5 cm thick, then the INNER dimensions are 7 cm x 9 cm x 11 cm

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 7x9 side, on the 7x11 side, or on the 9x11 side

If we place the base on the 7x9 side, then the cylinder will have height 11, and the maximum radius of the cylinder will be 3.5 (i.e., diameter of 7).
So, the volume of this cylinder will be (pi)(3.5²)(11), which equals (12.25)(11)(pi), which is a little bit more than 132pi

ASIDE: There's a nice trick for quickly calculating (in your head) the squares of numbers ending with 5 (e.g., 3.5²). See: https://www.gmatprepnow.com/module/gmat ... video/1024

If we place the base on the 7X11 side, then the cylinder will have height 9, and the maximum radius of the cylinder will be 3.5 (i.e., diameter of 7).
So, the volume of this cylinder will be (pi)(3.5²)(9), which equals (12.25)(9)(pi), which is a little bit more than 108pi

If we place the base on the 9x11 side, then the cylinder will have height 7, and the maximum radius of the cylinder will be 4.5 (i.e., diameter of 9).
So, the volume of this cylinder will be (pi)(4.5²)(7), which equals (20.25)(7)(pi), which is a little bit more than 140pi

So, the greatest possible volumeis a little bit more than 140pi, and this occurs when the radius is 4.5


Here's a similar practice question: https://gmatclub.com/forum/the-inside-d ... 28053.html

Answer: C

Cheers,
Brent

Solution:

Outer dimensions are 12 x 10 x 8 (l x b x h)

Uniform thickness = 1/2 cm

Thus inner dimension (imagine the dimensions guarding an inner storage space inside the cardboard box)

= 12-1 , 10-1, 9-1 respectively for inner length, inner breadth and inner height

= 11 x 9 x8

If you imagine the right circular inside this storage space, it can have the maximum volume, when

The diameter of the cylinder = inner width of the box

The diameter cannot be 11 as it would not fit in the box with the other two dimensions as 7 and 9 cm.

If the diameter is 7 which implies the width of the box is 7 and the other two dimensions are 9 and 11,

the volume in such a case would be pi * (3.5)*(3.5)*11

(To maximize the volume we take the larger value b/w 9 and 11 as height)

= pi * 12.25 * 11 = 134.75 pi

If the diameter is 9cm and the height is 7 cm (If the height was 11,the length of the box would be 7 and the canister would not fit in the box),the

volume of the canister would be pi * 20.25* 7

= 141.75 pi

Thus at width of the box = diameter of the canister =9cm,the volume is maximum

=>Radius for the maximized volume= 9/2 =4.5 cm (option c)

Hope this helps

Devmitra Sen(GMAT Quant Expert)
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Hi Bunuel unraveled
Quick question, what does outer radius imply here? I got the radius at 4.5 but interpreted outer radius to be 4.5+width and chose 5. Could you please explain this for me?

Intuitively, in order to maximise the volume of cylinder, one has to maximise the radius. As one of the dimension if 12cm less 1 cm, the radius will be 5.5 cm. However, the second highest dimension is 9 cm (10 -1 ). So, 9cm is limiting factor for the diameter and the radius will be 4.5 cm.

PS: I got this wrong on the first try for solving hastily.

elPatron434
As the box is of uniform thickness, the inside dimensions would be odd integer i.e. 7, 9, 11 cm. Hence, the maximum inside diameter would be a non-integer, implying 5 can't be the answer.



Coming to your query, if your answer is 4.5 then thickness(of canister) is already counted in that.
Why would you add 0.5 as thickness again, i can't understand.

I hope i understood your point correctly.
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Thanks for your help unraveled.

The question stem says that the width is 0.5 cm uniformly across. We have deducted 1cm from each dimension to discount the width.

Now, the maximum radius of the cyclinder in the inner box is 4.5 cm. The question asks us for the maximum outer radius. I assumed this to be inner radius + width. So, I added the width of 0.5 cm to it.

I hope I was able to elaborate on my query better. Thanks again

Posted from my mobile device

I think the red text is nothing but just an overkill. You overthought it i guess. 'maximum radius' and maximum outer radius' for cylinder(canister) both are same.
Also, the question does specifically mentioned thickness for box not the cylinder. For cylinder it can be anything but we are not concerned about it.

I can understand what you might have felt after knowing the answer, but that's just how it is.
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Understood! Thanks unraveled