Question

Find the volume of the largest circular cone that can beinscribed in a shpere of radius 3.

Answer 1

Answer:

$V =\dfrac{32}{3}\pi$

Step-by-step explanation:

given,

radius of sphere = 3

volume of cone:

$V = \dfrac{1}{3}\pi r^2h$

r is the radius of circular base

h is the height of the cone

here r = x and h = 3 + y

now, volume in term of x and y

$V = \dfrac{1}{3}\pi x^2(3+y)$

Applying Pythagoras theorem

x² + y² = 3²

$x = \sqrt{9-y^2}$

$V = \dfrac{1}{3}\pi ( \sqrt{9-y^2})^2(3+y)$

$V = \dfrac{1}{3}\pi ( 9-y^2)(3+y)$

$V = \dfrac{1}{3}\pi (27 + 9 y - 3 y^2-y^3)$

differentiating both side

$\dfrac{dV}{dy} =\dfrac{1}{3}\pi ( 9-6y- 3y^2)$

for maxima  $\dfrac{dV}{dy} = 0$

$\pi ( 3-2 y - y^2)=0$

 y² + 2 y - 3 = 0

(y+3)(y-1)=0

 y = 1,-3

y cannot be negative so, volume at y = 1

$V = \dfrac{1}{3}\pi (27 + 9 (1)- 3(1)^2-(1)^3)$

$V =\dfrac{32}{3}\pi$

Hence, the largest cone which can be inscribed in the spheres of the radius 3 has volume  $V =\dfrac{32}{3}\pi$