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Volume Problem

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by thp510 » Tue Nov 02, 2010 7:30 pm
A rectangular box has the dimensions 12x10x8 (inches). What is the largest possible volumne of a right cylinder that is placed inside the box?


ANSWER: The radius of the cylinder must be equal to half of the smaller of the 2 dimensions that form the box's bottom. So the ANS is 200pi (V=25pi * 8).

Question: Why would I only want to use the two smallest radius for the base of the volume and not the two largest radius? For instance, if I said the bottom of the cylinder has a radius of 6 and the height was 10, then the volume would be greater (V=360pi).
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by rajatdelhi » Tue Nov 02, 2010 8:18 pm
Volume of a right Cylinder = pi*r^2*h
Therefore, to have the greatest volume, we need to find the greatest possible radius.
Given the dimensions, the cylinder will have the greatest radius when the base is 12x10 and the height is 8.
Since the base is 12x10. The greatest radius possible will be (10/2) 5.
Making the volume = pi*25*8= 200pi
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by Geva@EconomistGMAT » Wed Nov 03, 2010 1:46 am
thp510 wrote:A rectangular box has the dimensions 12x10x8 (inches). What is the largest possible volumne of a right cylinder that is placed inside the box?


ANSWER: The radius of the cylinder must be equal to half of the smaller of the 2 dimensions that form the box's bottom. So the ANS is 200pi (V=25pi * 8).

Question: Why would I only want to use the two smallest radius for the base of the volume and not the two largest radius? For instance, if I said the bottom of the cylinder has a radius of 6 and the height was 10, then the volume would be greater (V=360pi).
If the height is 10, then the base is a rectangle 8*12. The 8 side of the rectangle limits the circle - the radius can't be 6, because a width of 8 cannot accommodate the diameter of 6*2=12. (think about it - the radius needs to be 6 in all directions from the center).
If the base were a square of 12 by 12, then the radius would be 6; but since it's a rectangle, the radius is limited by the smaller of the two dimensions of the bottom.
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