Among all right circular cones with a slant height of 27​, what are the dimensions​ (radius and​ height) that maximize the volume of the​ co

Question

Among all right circular cones with a slant height of 27​, what are the dimensions​ (radius and​ height) that maximize the volume of the​ cone? The slant height of a cone is the distance from the outer edge of the base to the vertex.

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Savannah 1 week 2021-11-23T16:23:29+00:00 1 Answer 0

Answers ( )

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    2021-11-23T16:25:25+00:00

    Answer:

    Radius = 22.04 unit,

    Height = 15.58 unit

    Step-by-step explanation:

    Since, the volume of cone,

    V=\frac{1}{3}\pi x^2 h

    Where,

    x = radius,

    h = height,

    Since, the slant height of a cone is,

    l=\sqrt{x^2 + h^2}

    l^2 = x^2 + h^2

    \implies h=\sqrt{l^2 - x^2}

    Here, l = 27.

    \implies h = \sqrt{27^2 - x^2}=\sqrt{729 - x^2}----(1)

    So, the volume would be,

    V=\frac{1}{3}\pi x^2 (\sqrt{729 - x^2})

    Differentiating with respect to x,

    \frac{dV}{dt}=\frac{1}{3}\pi (x^2 \times \frac{1}{2\sqrt{729 - x^2}}\times -2x + \sqrt{729 - x^2}\times 2x)

    =\frac{1}{3}\pi (\frac{-x^3+2x(729-x^2)}{\sqrt{729 - x^2}}

    =\frac{1}{3}\pi (\frac{-x^3+1458x-2x^3}{\sqrt{729 - x^2}})

    =\frac{1}{3}\pi (\frac{-3x^3+1458x}{\sqrt{729 - x^2}})

    =\pi (\frac{-x^3+486x}{\sqrt{729 - x^2}})

    For maxima or minima,

    \frac{dV}{dt}=0

    \implies -x^3 + 486x = 0

    x^3 = 486x

    x^2 = 486

    \implies x=\pm \sqrt{486}=\pm 22.04

    But side can not be negative,

    So, x = 22.04,

    Since,

    \frac{dV}{dx}|_{x=20}=297.91\text{Positive}

    While,

    \frac{dV}{dx}|_{x=23}=-219.70\text{Negative}

    Thus, by the first derivative test,

    V(x) is maximum at x = 22.04,

    Also, from equation (1),

    h = 15.58

    Hence, for maximising the volume, the radius and height of the cone would be 22.04 unit and 15.58 unit respectively.

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