Minimal Volume of a Cone Circumscribed about a Sphere

Calculus Tutorial

Minimal Volume of a Cone Circumscribed about a Sphere

Intro

Find the minimal volume and dimensions of a right circular cone circumscribed about a sphere of a given volume. To solve this problem we need to know

1) The formula for the volume of a sphere

V = \frac{4 {\pi} r^{2}}{3}

2) The formula for the volume of a cone

V = \frac{{\pi} r^{2} h} {3}

3) The radius of a sphere is perpendicular to a tangent line to the sphere.

4) Setting up ratios using similar triangles.

Sample Problem

Find the volume, radius, and height of a right circular cone of smallest volume about a sphere of volume 288\pi m^{3}.

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Solution

Let V_s represent the volume of the sphere, x represent the radius of the sphere and V_c, r and h represent the volume, radius and height of the right circular cone. I will solve the problem in general for any volume of a sphere.

1) Determine the radius of the sphere.

V_s=\frac{4{\pi}x^{3}}{3} \longrightarrow x^{3}=\frac{3V_s}{4\pi} \longrightarrow x=\sqrt[3]{\frac{3V_s}{4\pi}}

2) Set up similar triangles using the geometry of the problem and represent the volume of a cone a function of h.

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\frac{r} {\sqrt{h^2 + r^2}} = \frac{x} {h - x}

Squaring each side gives

\frac{r^2} {h^2 + r^2} = \frac{x^2} {\left(h - x \right)^2}

Now solve for r^2 in terms of h.

r^2(h-x)^2 = x^2h^2 + x^2r^2, cross multiplying

r^2(h-x)^2 -x^2r^2 = x^2h^2, move all terms containing r^2 to the right

r^2((h-x)^2 -x^2) = x^2h^2, factor out r^2

r^2(h^2 - 2xh) = x^2h^2, foil, combine like terms then divide each side by h^2 - 2xh

r^2 = \frac{x^2h^2} {h^2 - 2xh} \longrightarrow r^2 = \frac{x^2h} {h - 2x}

Now substitute this value for r^2 into formula for the volume of the cone

V_c = \frac{\pi r^2h} {3} \longrightarrow \frac{\pi x^2 h^2} {3(h - 2x)} \longrightarrow V_c = \frac{\pi x^2}{3} \cdot \frac{h^2} {h-2x},\quad D_{V_c} = (2x, \infty)

I have parted V_c to make taking the derivative easier.

3) Take the derivative of V_c with respect to h using the quotient rule.

V_{c}^{'} = \frac{\pi x^2}{3} \left[\frac{(h - 2x)2h - h^2(1)}{(h-2x)^2}\right]

V_{c}^{'} = \frac{\pi x^2}{3}\left[\frac{h^2 - 4xh}{(h-2x)^2}\right]

4) Set V_{c}^{'} = 0 and solve for h.

\frac{\pi x^2}{3}\left[\frac{h^2 - 4xh}{(h-2x)^2}\right] = 0 \to \frac{h^2 - 4xh}{(h-2x)^2} = 0

h^2 - 4xh = 0 \to h(h-4x) = 0 \to h = 0 \quad or \quad h-4x = 0 \to h = 4x

Note that the derivative does not exist at h=2x but 2x and 0 is not in the domain so the only possible value for h is 4x.

Now solve for the radius.

r^2 = \frac{x^2h} {h - 2x} \to r = \sqrt{\frac{x^2h} {h - 2x}}

h = 4x \to r = \sqrt{\frac{x^2 \cdot 4x} {(4x - 2x)}} = \sqrt{\frac{4 x^3} {2x}} = x\sqrt{2}

5) Solve for the volume of the cone representing all values in terms of V_s.

V_c = \frac{\pi (x\sqrt{2})^2(4x)}{3} = \frac{8x^3 \pi}{3} = \frac{8\pi}{3}(\frac{3V_s}{4 \pi}) = 2V_s

r = x\sqrt{2} = \left(\sqrt[3]{\frac{3V_s}{4 \pi}}\right)\left(\sqrt{2}\right) = \sqrt[6]{\frac{9V_s^2} {16\pi^2}}\sqrt[6]{8} \to r = \sqrt[6]{\frac{9V_s^2} {2 \pi^2}}

h = 4x = 4\left(\sqrt[3]{\frac{3V_s}{4 \pi}}\right)

Substituting in the numerical value of V_s we have

V_c = 2(288 \pi) = 576 \pi \approxeq 1809.5574m^3

r = \sqrt[6]{\frac{9(288\pi)^2} {2 \pi^2}} = \sqrt[6]{373248}} \approxeq 8.4853m

h = 4\sqrt[3]{\frac{3(288\pi)} {4 \pi}} = 24m



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I am a graduate of Cleveland State University. After earning my Bachelors of Electrical Engineering I went on to earn a Master’s of Science in Electrical Engineering. Also I am a graduate of Lorain County Community College (LCCC) earning an Associates of Science Pre-Professional Engineering and ...
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