Find-maximum-volume-of-a-cylinder-within-a-unit-cube-whose-axis-is-along-a-diagonal-of-the-cube-

Question Number 138388 by ajfour last updated on 13/Apr/21

F i n d m a x i m u m v o l u m e o f a

c y l i n d e r w i t h i n a u n i t c u b e

w h o s e a x i s i s a l o n g a d i a g o n a l

o f t h e c u b e .

Answered by mr W last updated on 13/Apr/21

r = r a d i u s o f c y l i n d e r

l = l e n g t h o f c y l i n d e r

a = e d g e l e n g t h o f c u b e = 1

l = \sqrt{3} a - 2 \sqrt{2} r

V = π r^{2} l = π r^{2} (\sqrt{3} a - 2 \sqrt{2} r)

\frac{d V}{d r} = π (2 \sqrt{3} a r - 6 \sqrt{2} r^{2}) = 0

\Rightarrow r_{m} = \frac{a}{\sqrt{6}}

V_{m a x} = π \times \frac{a^{2}}{6} (\sqrt{3} a - \frac{2 \sqrt{2} a}{\sqrt{6}}) = \frac{\sqrt{3} π a^{3}}{18}

Commented by mr W last updated on 13/Apr/21

f i n e! {i t}^{'} s m o r e e l e g a n t u s i n g v e c t o r s .

Commented by ajfour last updated on 13/Apr/21

y e s s i r, t h a n k y o u, u n d e r s t o o d .

\cos ϕ = \frac{(i + j + k)}{\sqrt{3}} \cdot \frac{(i + j)}{\sqrt{2}} = \frac{\sqrt{2}}{\sqrt{3}}

\tan ϕ = \frac{r}{b} = \frac{1}{\sqrt{2}} \Rightarrow b = r \sqrt{2}

2 b + l = a \sqrt{3}

2 r \sqrt{2} + l = a \sqrt{3}

l = a \sqrt{3} - 2 r \sqrt{2}

\dots .

\dots .

r_{m} = \frac{a}{\sqrt{6}}, V_{m a x} = \frac{\sqrt{3} π a^{3}}{18}

Commented by mr W last updated on 14/Apr/21

c a n y o u p r o v e Q 138519 u s i n g v e c t o r

m e t h o d ? i f o u n d t h a t b y a c c i d e n t, b u t

h a v e n o p r o o f y e t .

Answered by mr W last updated on 14/Apr/21

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