# a right circular cone is inserted in a sphere of radius(R). if the height of the cone is R+x, express the volume in term of x and R.?

please i really help with this question.

Update:

if R is fixed and x is allowed to vary, find the maximum value of the volume of the cone.

Relevance

let r = the radius of the base of the cone , we also have:

R + x = height cone

V = (1/3) * pi * r^2 * (R + x) => find r^2 in terms of R and x:

r^2 = R^2 - x^2, so:

V = (1/3) * pi * (R^2 - x^2) * (R + x)

V = (1/3) * pi * [R^3 + R^2x - x^2R - x^3] => derivative with respect to x

dV/dx = (1/3) * pi * [R^2 - 2xR - 3x^2] => set to zero

R^2 - 2xR - 3x^2 = 0

(R + x)(R - 3x) = 0

x = -R , R/3 => reject the negative root:

x = R/3 , use the 2nd derivative to determine the nature, i.e: Max or Min:

V" = (1/3) * pi * [-2R - 6] < 0 , hence it's a local/relative maximum, then:

r^2 = R^2 - x^2 => sub for x = R/3:

r^2 = R^2 - (R/3)^2

r^2 = R^2 - R^2/9 = (8R^2)/9

R + x = R + R/3 = (4R)/3

then the maximum volume would be:

V = (1/3) * pi * (R^2 - x^2) * (R + x) => sub. for R^2 - x^2 = (8R^2)/9 and for R + x = (4R)/3:

V = (1/3) * pi * (8R^2)/9 * (4R)/3

V = 32/81 * pi * R^3

I hope this helps.

• Volume of a cone=1/3*pi*base radius^2*height

So the area of the right circular cone=1/3*R^2*(R+x)=R^3/3+R^2*x/3

• Height of the cone, h = R+x

i.e. base is at a distance x from the centre of the sphere

Radius of teh base at distance x fromthe center, r=√(R²−x²)

Hence volume of the cone =(1/3)πr²h=(1/3)π{√(R²−x²)}²(R+x)

=(1/3)π(R²−x²)(R+x)=(1/3)π(R²+R²x−Rx²−x²)

×√+−×√+−×√