Cylinder

[sudhir3127]

1. A right circular cylinder is to be made out of a metal sheet such that the sum of its height and radius doesnt exceed 9 cm can have an area of maximum.

Answers after some discussion.!!

[ramyaravindran]

The Surface Area of the cylinder is given by the formula 2*(pi)*r^2 + 2*(pi)*r*h

This can be written as 2*pi*r*(r+h)

We are given that r+h => 9

So the surface area is 2*pi*r*9 => 18*pi*r...this can be maximized when r=8 and h =1

[nandistl]

assuming r and h are integers and r + h <= 9

max area ca be be 144 Pi. (2* pi * r *(r + h))

[sudhir3127]

Thanks a lot all for the replies..

The answer is 108 pi

here it goes...

the answer is 108Pi.. when we take r=6 and h =2 and we need to calculate the volume .

to arrive @ 6 and 2 we need to know the rule of ratios and proportion...

For a^2b^3c^4 to be maximum ,a+b+c should be constant and a,b,c should be in the ratio of 1:2:3

usind this we know that the radius is 6 and height is 2.

Volume of cylinder = pi*r^2*h = 108 pi.

PS: Ian/ Stuart/Ron .. i would like to hear from you all as to whether the approach is right.. or there is better way to do it.It would be helpful if you put forth ur viewpoint on this,

Thanks !!

[santa_dem]

Thanks a lot all for the replies..

The answer is 108 pi

here it goes...

the answer is 108Pi.. when we take r=6 and h =2 and we need to calculate the volume .

to arrive @ 6 and 2 we need to know the rule of ratios and proportion...

For a^2b^3c^4 to be maximum ,a+b+c should be constant and a,b,c should be in the ratio of 1:2:3

usind this we know that the radius is 6 and height is 2.

Volume of cylinder = pi*r^2*h = 108 pi.

PS: Ian/ Stuart/Ron .. i would like to hear from you all as to whether the approach is right.. or there is better way to do it.It would be helpful if you put forth ur viewpoint on this,

Thanks !!

You were talking about area in your first post. How did it get to volume?

[Ian Stewart]

If the question is to maximize surface area, then, as is pointed out in solutions above, we find that SA = 18*Pi*r, so the surface area will be largest when r is largest. The maximum surface area is larger than 144*Pi, however, since r does not need to be an integer. Since r can take any value less than 9, the surface area could take any value less than 162*Pi.

If the question is to maximize volume, Sudhir has found the correct radius (6), though the height should be 3, not 2. I don't follow the approach, however. This is a problem that would normally be solved using calculus, though if you have answer choices, it is not difficult to test them here. If you know calculus (which you don't need for the GMAT!) you can solve as follows:

V = Pi*r^2*h = Pi*r^2*(9-r) = 9*Pi*r^2 - Pi*r^3
V' = 18*Pi*r - 3*Pi*r^2

Set V' = 0 to find maxima and minima:

0 = 18*Pi*r - 3*Pi*r^2
0 = r*(18*Pi - 3*Pi*r)

So we have a minimum value when r = 0, and a maximum value when r = 6. Plugging r = 6 in to find V, we see that V = 108*Pi.

If you don't know calculus, the above probably won't make much sense- it's definitely not something tested on the GMAT.

[santa_dem]

I don't think this is a GMAT problem, but it's good for practicing area/volume of figures.

[sudhir3127]

Sorrie for the typo..i infact got radius as 6 and height 3. and the volume was 108pi.

its a CAT 2004 ( India - IIM) question.

thanks Ian for all ur help. highly appreciated..