Latest Elitmus Aptitude Question SOLUTION: A right circular cone is of height 1m and radius 6m. what is the maximum volume of cylinder that can be cut from this cone??? a) π 2√3 b) 16π/3 c) 9π/2 d) dnt remember

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(#M40038393) ELITMUS Numerical Ability question area and volume 17-aug-14 Keep an EYE

A right circular cone is of height 1m and radius 6m. what is the maximum volume of cylinder that can be cut from this cone???
a) π 2√3
b) 16π/3
c) 9π/2
d) dnt remember

Asked In Elitmus ankita soni (9 years ago)

Unsolved Read Solution (16)

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RAKESH  (9 years ago)

b) 16π/3

maxm volume of cylinder can be cut for
r = 4 m, h = 1/3 m [use concept of maxima & minima]

vol = pi*4^2*(1/3) = 16*pi/3

Is this solution Helpfull? Yes (30) | No (17)  

arjun kumar jha  (9 years ago)

radius of cylinder will be 3...and height will be reduced to 1/2...
so volume will be 9 pie/2

Is this solution Helpfull? Yes (17) | No (7)  

Sujoy Das  (9 years ago)

http://www.mathalino.com/reviewer/differential-calculus/cylinder-maximum-volume-and-maximum-lateral-area-inscribed-cone

Is this solution Helpfull? Yes (12) | No (2)  

raj  (9 years ago)

answer 9pi/2

height will be half as well radius will also decreased by 50% and now taking radius as 6/2=3 and height as 1/2 calculate pi*3^2*1/2=9pi/2

Is this solution Helpfull? Yes (4) | No (3)  

arjun kumar jha  (9 years ago)

answer is 16n/3....
explanation.
we will have to use differentiation technique to get result...consider a cylinder of radius x and height 1-y...and y will be the height of rest of the cone coz total height is 1....to get the volume of cylinder in terms of y..follow simly of trngle...we will reach...x=6*y....
lets volume of cylinder be v=pie*r*r*h...this implies pie*36*y^2*(1-y)..differentiate v wrt y.. and to get max volume it will be made eyual to zero.
we will get y=2/3..and x=4...required volume will be pie*4*4*(1-2/3) i.e 16pie/3

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raushan kumar  (9 years ago)

1-h/1 = r/6 h = hight of cylinder , r = radius of cone.(using ~ triangle concept)

h = r-6/6

dV/dr=0, v= 1/3*pi*r^2*(r-6/6) (V of cone)
r=4 , h= 1/3.

then volume of cyl. = pi*r^2*h = Pi*16/3.

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raushan kumar  (9 years ago)

1-h/1 = r/6 h = hight of cylinder , r = radius of cone.(using ~ triangle concept)

h = 1-r/6

dV/dr=0, v= 1/3*pi*r^2*(1-r/6) (V of cone)
r=4 , h= 1/3.

then volume of cyl. = pi*r^2*h = Pi*16/3.

Is this solution Helpfull? Yes (2) | No (1)  

amit kumar  (9 years ago)

damm sure ans is 16pi/3

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Sujoy Das  (9 years ago)

Let the radius of cylinder=x;height of cylinder=h;
y is the length from vertex of the cone to the top surface of the cylinder
y/x=6/1;
Volume of cylinder=pi*x^2*h;
h=6-(6*x);
Volume=pi*x^2*(6-(6*x))
Now to get the maximum volume we have to differentiate Volume with respect to x
6*pi[2x-3x^2]=0;
x=2/3;
Therefore h=6-6*2/3;
h=2;
Volume=pi*(2/3)^2*2;
Volume=8pi/9;

Is this solution Helpfull? Yes (1) | No (3)  

nidhi gupta  (9 years ago)

how? can u explain in details?

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ashish  (9 years ago)

64pie/3

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Sujoy Das  (9 years ago)

4th) option is 8pi/9 and this is the answer.I am 100% sure

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sandeep kiran  (9 years ago)

c)9pie/2
take radius of cylinder x and by similarity find the height of cylinder in terms of radius and height of cone.
h1=h(r-x)/r
where h1 is height of cylinder
h height of cone
r raidus of cone
and x radius of cylinder.
Now v=pie*r^2h1,put value n diff in term of x...
use maxima n minima concept to conform the maximum volume at x=r/2.
also h1 =h/2.
put value get d result.

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ranjitkumar  (9 years ago)

assume six possible cylender by deviding redis and hight in equally ratio
where height =(1/6,/6,/6,/6,/6,/6) and ratio= (1,1,1,1,1,1)
then volum of cylender are
pi*25*1/6=4.166pi
pi*16*(1/6)*2=5.34pi
pi*9*(1/6)*3=4.5pi
pi*4*(1/6)*4=2.7pi
pi*1*(1/6)*5=0.84pi

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ranjitkumar  (9 years ago)

radious........@ratio

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Mithun Kumar Kesari  (9 years ago)

thanks @sujoy das

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