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Question Number 118069 by Lordose last updated on 15/Oct/20

1 .) i) A right circular cone is circumscribed

about a sphere of radius (r) . If d is the

distance from the center of the sphere

to the vertex of the cone, show that the

volume of the cone, V = \frac{π r^{2} {(r + d)}^{2}}{3 (d - r)} .

ii)

F ind the vertical angle of the cone when

{it}^{'} s volume is minimum .

Answered by 1549442205PVT last updated on 15/Oct/20

i) Altitude of the cone equalto h = d + r

Denote by φ the at the vertex of the

cone \Rightarrow \tan \frac{φ}{2} = \frac{r}{\sqrt{d^{2} - r^{2}}},

The radius of base circle equal to

R = htan \frac{φ}{2} = \frac{(d + r) r}{\sqrt{d^{2} - r^{2}}}

\Rightarrow V = \frac{1}{3} Bh = \frac{1}{3} . π R^{2} h = \frac{π r^{2} {(d + r)}^{3}}{3 (d^{2} - r^{2})}

V = \frac{π r^{2} {(d + r)}^{2}}{3 (d - r)} (q . e . d)

ii) There are two possible possbility ocurr

a) The case r = constant

Put x = d . we need determine d when

V receive smallest value . Since r is

constant, so Vhas greatest value if and

only if f (x) = \frac{{(x + r)}^{2}}{x - r}, x \in (0, 2 r) is smallest

f^{'} (x) = \frac{2 (x + r) (x - r) - {(x + r)}^{2}}{{(x - r)}^{2}} = 0

\Leftrightarrow x^{2} - 2 xr - {3 r}^{2} = 0 Δ^{'} = r^{2} + {3 r}^{2} = {(2 r)}^{2}

\Rightarrow x = 3 r \notin (0, 2 r) . Thus if r = constant then

V {can}^{'} t receive smallest

b) The d = constant, put r = x . Then V

receives smallest value if and only if

f (x) = \frac{x^{2} {(d + x)}^{2}}{d - x} receives smallestvalue

f^{'} (x) = \frac{2 x (d + x) (2 x + d) + x^{2} {(d + x)}^{2}}{{(d - x)}^{2}} = 0

\Leftrightarrow {4 x}^{3} + {6 dx}^{2} + {2 d}^{2} x + x^{4} + {2 dx}^{3} + d^{2} x^{2} = 0

It is easy to see this equation has no

roots in interval (0, d)

Thus, V {can}^{'} t receives smallest value

The problem has no solution