What-should-the-dimensions-be-of-a-cylinder-so-that-for-a-given-volume-v-its-total-surface-S-minimum-

Question Number 124924 by bemath last updated on 07/Dec/20

W h a t s h o u l d t h e d i m e n s i o n s b e

o f a c y l i n d e r s o t h a t f o r a g i v e n v o l u m e

v i t s t o t a l s u r f a c e S m i n i m u m ?

Commented by liberty last updated on 07/Dec/20

D e n o t i n g b y r t h e r a d i u s o f t h e b a s e o f t h e

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

S = 2 π r^{2} + 2 π r h a n d t h e v o l u m e i s g i v e n

v = π r^{2} h; w h e n c e h = \frac{v}{π r^{2}}

s u b s t i t u t i n g t h i s e x p r e s s i o n o f h i n t o t h e

f o r m u l a f o r S, g i v e S = 2 (π r^{2} + \frac{v}{r}); w h e r e 0 < r < \infty

\frac{d S}{d r} = 2 (2 π r - \frac{v}{r^{2}}) = 0 \Rightarrow r_{1} = \sqrt[3]{\frac{v}{2 π}}

{(\frac{d^{2} S}{{d r}^{2}})}_{r = r_{1}} = 2 {(2 π + \frac{2 v}{r^{3}})}_{r = r_{1}} > 0

T h u s a t t h e p o i n t r = r_{1} t h e f u n c t i o n S h a s

a m i n i m u m . N o t i c i n g t h a t \lim_{r \to 0} S = \infty a n d

\lim_{r \to \infty} S = \infty, w e a r r i v e a t c o n c l u s i o n t h a t

a t r = r_{1} = \sqrt[3]{\frac{v}{2 π}} t h e n h = \frac{v}{π r^{2}} = 2 \sqrt[3]{\frac{v}{2 π}} = 2 r

Answered by som(math1967) last updated on 07/Dec/20

let radius = r height = h

v = π r^{2} h \Rightarrow h = \frac{v}{π r^{2}}

S = 2 π rh + 2 π r^{2}

S = \frac{2 v}{r} + 2 π r^{2}

\frac{dS}{dr} = - \frac{2 v}{r^{2}} + 4 π r

\frac{d^{2} S}{{dr}^{2}} = \frac{4 v}{r^{3}} + 4 π

∴ \frac{d^{2} S}{{dr}^{2}} > 0 ∴ S minimum for

\frac{dS}{dr} = 0

∴ - \frac{2 v}{r^{2}} + 4 π r = 0

4 π r^{3} = 2 v

2 π r^{3} = π r^{2} h

2 r = h

∴ S is minimum when

Height is eqal to diameter

or r : h = 1 : 2