Rectangular Crate

Sorry I can't draw a cube here, but try and imagine a simple cube to start with, with length = 8 , height = 10 and depth = 12.

The radius of the cylinder will be half the dimension it rests against (e.g. is the cube is resting with the length of the cube as its base, then radisu = 8/2 = 4)

The cylinder can be placed in 3 different configurations, so you need to test each:

Config 1:

Height = 10, radius = 4, area = 160pi

Config2:

Height = 8, radius = 6, area = 288pi

config 3:

Height = 12, radius = 5, area = 300pi

As we can see, radius = 5 gives us the max. volume of the cylinder.

kmittal82 wrote:Sorry I can't draw a cube here, but try and imagine a simple cube to start with, with length = 8 , height = 10 and depth = 12.

The radius of the cylinder will be half the dimension it rests against (e.g. is the cube is resting with the length of the cube as its base, then radisu = 8/2 = 4)

The cylinder can be placed in 3 different configurations, so you need to test each:

Config 1:

Height = 10, radius = 4, area = 160pi

Config2:

Height = 8, radius = 6, area = 288pi

config 3:

Height = 12, radius = 5, area = 300pi

As we can see, radius = 5 gives us the max. volume of the cylinder.

Hi,

your answer is correct, but you've given impossible numbers.

We have 3 potential sides for our radius:

8*10, 8*12 and 10*12. However, the length of the radius is limited by the shorter of the two dimensions, not the longer.

So, to get radius 5, you have to put the base of the cylinder on the 10*12 side and you'll have a height of 8, not 12.

Numbers aside, here's a conceptual way to approach this rather than working out all the actual volumes:

the formula for the volume of a cylinder is (pi)(r^2)(h). Since the radius is squared, it will have a much bigger impact on the volume than will the height.

So, to maximize volume we'll almost always want to maximize the radius - since the biggest radius we can get in this box is 5, that's the right answer to the question.

You may reply "well, I can come up with really weird dimensions for which it would make sense to maximize the height instead" (e.g. 1*2*1000, we'd be better off with radius 1 and height 1000, since so much of the box gets wasted if we use the 2*1000 side to create a radius of 2). However, since the GMAT wants to reward you for understanding concepts rather than just test your ability to crunch numbers, it's extremely unlikely that you'd see those dimensions on test day.

Hi Stuart, your answer reminds me....

that if we are really short of time... say 2 mins and 5 questions....

then its best... to have an educated guess.....


sides are 8 10 and 12 , so the possible radii can be 4, 5,6 choose between A, B and C and move ahead...

if u have more time...


8-10 as base=> 12 *16pi
10-12 as base=> 8 *25pi --> your answer derived from here... radius is 5 -->B
8-12 as base=> 10*16pi

Stuart Kovinsky wrote:

kmittal82 wrote:Sorry I can't draw a cube here, but try and imagine a simple cube to start with, with length = 8 , height = 10 and depth = 12.

The radius of the cylinder will be half the dimension it rests against (e.g. is the cube is resting with the length of the cube as its base, then radisu = 8/2 = 4)

The cylinder can be placed in 3 different configurations, so you need to test each:

Config 1:

Height = 10, radius = 4, area = 160pi

Config2:

Height = 8, radius = 6, area = 288pi

config 3:

Height = 12, radius = 5, area = 300pi

As we can see, radius = 5 gives us the max. volume of the cylinder.

Hi,

your answer is correct, but you've given impossible numbers.

We have 3 potential sides for our radius:

8*10, 8*12 and 10*12. However, the length of the radius is limited by the shorter of the two dimensions, not the longer.

So, to get radius 5, you have to put the base of the cylinder on the 10*12 side and you'll have a height of 8, not 12.

Numbers aside, here's a conceptual way to approach this rather than working out all the actual volumes:

the formula for the volume of a cylinder is (pi)(r^2)(h). Since the radius is squared, it will have a much bigger impact on the volume than will the height.

So, to maximize volume we'll almost always want to maximize the radius - since the biggest radius we can get in this box is 5, that's the right answer to the question.

You may reply "well, I can come up with really weird dimensions for which it would make sense to maximize the height instead" (e.g. 1*2*1000, we'd be better off with radius 1 and height 1000, since so much of the box gets wasted if we use the 2*1000 side to create a radius of 2). However, since the GMAT wants to reward you for understanding concepts rather than just test your ability to crunch numbers, it's extremely unlikely that you'd see those dimensions on test day.

Thanks for correcting me there Stuart, teaches me to try and solve something under a minute!

arora007 wrote:Hi Stuart, your answer reminds me....

that if we are really short of time... say 2 mins and 5 questions....

then its best... to have an educated guess.....


sides are 8 10 and 12 , so the possible radii can be 4, 5,6 choose between A, B and C and move ahead...

Hi,

even better: if the sides are 8, 10 and 12 then the biggest radii we can make are 4 and 5 (remember, the smaller dimension limits the radius, not the bigger one) - so you have a very quick 50/50 shot.