Show that the altitude of the right circle cone of maximum cured surface which can inscribed in a sphere of radius r is $\frac{4 r}{3}$ .

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Solution

$A s p h e r e o f r a d i u s (r) i s g i v e n .$

$L e t R a n d h b e t h e r a d i u s a n d t h e h e i g h t o f t h e c o n e r e s p e c t i v e l y a n d t h e v o l u m e o f t h e c o n e i s V = \frac{1}{3} π R ² h$

$f r o m r i g h t a n g l e Δ B C D,$

$B C^{2} = r^{2} - R^{2}$

$h = r + B C$

$V = \frac{1}{3} π R^{2} r + \sqrt{(r^{2} - R^{2})}$

$= \frac{1}{3} π R^{2} r + \frac{1}{3} π R^{2} \sqrt{(r^{2} - R^{2})}$

$n o w,$

$\frac{d V}{d R} = \frac{2}{3} π R r + \frac{2}{3} π R \sqrt{(r ² - R ²)} + \frac{R ²}{3} * \frac{(- 2 R)}{2 \sqrt{r ² - R ²}}$

$= \frac{2}{3} π R r + \frac{2}{3} 2 / 3 \sqrt{(r ² - R ²)} - \frac{π R ³}{3} \sqrt{(r ² - R ²)}$

$= 2 / 3 π R r + \frac{2 Π R (r^{2} - R^{2}) - Π R^{3}}{3 \sqrt{(r^{2} - R^{2})}}$

$= 2 / 3 π R r + \frac{2 Π R r^{2} - 3 Π R^{3}}{3 \sqrt{(r^{2} - R^{2})}}$

$n o w, \frac{d V}{d R} = 0$

$\frac{2}{3} π R r = \frac{3 π R ³ - 2 π R r ²}{3 \sqrt{(r ² - R ²}}$

$2 r \sqrt{(r ² - R ²)} = 3 R ² - 2 r ²$

$n o w o n s q u a r i n g b o t h s i d e,$

$4 r ² (r ² - R ²) = 9 R ⁴ + 4 r ⁴ - 12 R ² r ²$

$4 r ⁴ - 4 r ² R ² - 4 r ⁴ + 12 r ² R ² = 9 R ⁴$

$8 r ² R ² = 9 R ⁴$

$R ² = \frac{8 r ²}{9}$

$w h e n, R ² = 8 r ² / 9,$

$(2 Π r^{2} - 9 Π R^{2})$

$t h e r e f o r e, v o l u m e i s m a x i m u m w h e n$

$R ² = \frac{8 r ²}{9}$

$s o, h e i g h t o f t h e c o n e h = r + \sqrt{(r ² - R ²)}$

$h = r + \sqrt{r^{2} - \frac{{8 r}^{2}}{9}}$

$h = r + \sqrt{\frac{r^{2}}{9}}$

$h = r + \frac{r}{3}$

$h = 4 r / 3$

$H e n c e, i t c a n b e s e e n t h a t t h e a l t i t u d e o f t h e r i g h t c i r c u l a r c o n e o f m a x i m u m v o l u m e t h a t c a n b e i n s c r i b e d i n a s p h e r e o f r a d i u s r i s \frac{4 r}{3} .$

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