find the circular cylinder of greatest volume that can be inscribed in a right circular cone with altitude?

Question

20 cm and diameter of base 10 cm.

(optimization problem)
calculus

answer in detail plz

Answer 1

If you slice the cone you have an altitude of 20 and a diameter of 10 so drop a perpendicular to get a right triangle of height 20 and base 5. If you look at the right triangle on the right side, the hypotenuse can be viewed as a line with equation y=-4x+20 because you have the points (5,0) from the radius and (0,20) from the height. So for any x between 0 and 5, the corresponding height will be -4x+20. Maximizing the area of this rectangle also maximizes the volume of the circular cylinder. The area is A=x* (-4x+20). When you find A' and set it to zero and solve, you get x=2.5 cm.

Answer 2

Look at the triangle cross-section of a cone. You get rectangular triangle of base 5 cm (half of base diameter) and hight 20 cm (altitude of the cone). For inscribed cylinder with base radius r and cylinder hight h: h/(5-r) = 20/5 = 4. Therefore h = 4(5-r). Cylinder volume is V = r^2*pi*h = 20*r^2*pi - 4*r^3*pi. First derivative dV/dr = 40*r*pi - 12*r^2*pi = 0 gives r = 20/6 cm. Second derivative at this point = -40*pi which prove that V at this point is maximum. Therefore Vmax = 2000*pi/27.

Answer 3

Let the base radius x The hight of the cylinder y=-2x+6 Vol= pi x^2(-2x+6) V=-2 x^3+6x^2 dV/dx= -6x^2+12x=0 for the extreem vol. X=0 for the min x=2 for the max vol Vmax=8 pi

Answer 4

Volume of Cylinder = Pi*R**2*H
H=20-R*4
V=Pi*20*R**2-Pi*R**3*4
dV/dR=Pi*20*2*R-Pi*3*R**2*4
dV/dR=Pi*4*R*(10-3*R)
dV/dR has a zero at R=10/3
therefore the largest cylinder has
a radius of 3.3333...
and from H=20-R*4
a height of 6.6666...