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

• 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
• 9 years agoHelpfull: Yes(30) No(17)
• radius of cylinder will be 3...and height will be reduced to 1/2...
  so volume will be 9 pie/2
• 9 years agoHelpfull: Yes(17) No(7)
• http://www.mathalino.com/reviewer/differential-calculus/cylinder-maximum-volume-and-maximum-lateral-area-inscribed-cone
  
• 9 years agoHelpfull: Yes(12) No(2)
• 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
• 9 years agoHelpfull: Yes(4) No(3)
• 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
• 9 years agoHelpfull: Yes(4) No(10)
• 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.
  
  
• 9 years agoHelpfull: Yes(4) No(3)
• 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.
• 9 years agoHelpfull: Yes(2) No(1)
• damm sure ans is 16pi/3
  
• 9 years agoHelpfull: Yes(2) No(1)
• 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;
• 9 years agoHelpfull: Yes(1) No(3)
• how? can u explain in details?
• 9 years agoHelpfull: Yes(0) No(0)
• 64pie/3
• 9 years agoHelpfull: Yes(0) No(2)
• 4th) option is 8pi/9 and this is the answer.I am 100% sure
  
• 9 years agoHelpfull: Yes(0) No(2)
• 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.
• 9 years agoHelpfull: Yes(0) No(0)
• 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
  
• 9 years agoHelpfull: Yes(0) No(1)
• radious........@ratio
• 9 years agoHelpfull: Yes(0) No(0)
• thanks @sujoy das
  
• 9 years agoHelpfull: Yes(0) No(1)