Answer:
[tex]V =\dfrac{32}{3}\pi[/tex]
Step-by-step explanation:
given,
radius of sphere = 3
volume of cone:
[tex]V = \dfrac{1}{3}\pi r^2h[/tex]
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
[tex]V = \dfrac{1}{3}\pi x^2(3+y)[/tex]
Applying Pythagoras theorem
x² + y² = 3²
[tex]x = \sqrt{9-y^2}[/tex]
[tex]V = \dfrac{1}{3}\pi ( \sqrt{9-y^2})^2(3+y)[/tex]
[tex]V = \dfrac{1}{3}\pi ( 9-y^2)(3+y)[/tex]
[tex]V = \dfrac{1}{3}\pi (27 + 9 y - 3 y^2-y^3)[/tex]
differentiating both side
[tex]\dfrac{dV}{dy} =\dfrac{1}{3}\pi ( 9-6y- 3y^2)[/tex]
for maxima [tex]\dfrac{dV}{dy} = 0 [/tex]
[tex]\pi ( 3-2 y - y^2)=0[/tex]
y² + 2 y - 3 = 0
(y+3)(y-1)=0
y = 1,-3
y cannot be negative so, volume at y = 1
[tex]V = \dfrac{1}{3}\pi (27 + 9 (1)- 3(1)^2-(1)^3)[/tex]
[tex]V =\dfrac{32}{3}\pi[/tex]
Hence, the largest cone which can be inscribed in the spheres of the radius 3 has volume [tex]V =\dfrac{32}{3}\pi[/tex]
