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.

3 Answers

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  • iceman
    Lv 7
    5 years ago
    Favorite Answer

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

    R = radius of sphere

    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.

  • 5 years ago

    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

  • Raj K
    Lv 7
    5 years ago

    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)

    ×√+−×√+−×√

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