Question

The maximum volume (in cu. M) of the right circular cone having slant height 3m is :

Answer 1

Answer:

The maximum volume (in cu. M) of the right circular cone having slant height 3m is : 2π√3 m³

Step-by-step explanation:

Height of the cone = Lcosθ

Radius of the cone = L Sinθ

Volume of cone = (1/3)πr²h

=(1/3)πL²Sin²θLcosθ

=(1/3)πL³Sin²θcosθ

L = 3 m

= (1/3)π3³Sin²θcosθ

= 9πSin²θcosθ m³

To get maximum value

We need to differentiate it and we find

Sinθ = √(2/3)   then Cosθ = 1/√3

putting these values

Volume = 9π(2/3)/√3  = 6π/√3 m³  = 2π√3 m³

Answer 2

Answer:

2√3 π

Step-by-step explanation:

Given The maximum volume (in cu. M) of the right circular cone having slant height 3m is

Volume of a circular cone V = 1/3πr^2h

We know that l^2 = r^2 + h^2

                  3^2 = h^2 + r^2

                  r^2 = 9 - h^2

V = 1/3 π (9 - h^2)h

dv/dh = 0

dv/dh  =  π/3(9 - x^2) + hπ/3(0 - 2h)

π/3(9- h^2) + hπ/3(0 - 2h) = 0

9 - h^2 - 2h^2 = 0

9 - 3h^2 = 0

3 h^2 = 9

h = √3

Vmax = 1/3 π(9 - 3)(√3)

V max = 2√3 π c mtr