I-want-to-make-cyllinders-of-maximum-volume-V-max-with-minimum-surface-area-S-min-because-they-are-more-profitable-for-me-I-need-the-help-of-mathematicians-of-this-forum-

Question Number 3996 by Rasheed Soomro last updated on 26/Dec/15

I w a n t t o m a k e c y l l i n d e r s o f m a x i m u m

v o l u m e V_{\max} w i t h m i n i m u m s u r f a c e a r e a

S_{\min}, b e c a u s e t h e y a r e m o r e p r o f i t a b l e f o r m e .

I n e e d t h e h e l p o f m a t h e m a t i c i a n s o f t h i s

f o r u m .

Commented by Filup last updated on 26/Dec/15

V = π r^{2} h

S = 2 π r^{2} + 2 π r h

V_{m a x} when V^{'} = 0

\frac{d V}{d r} = 2 π r h o r \frac{d V}{d h} = π r^{2}

S_{m i n} when S^{'} = 0

\frac{d S}{d r} = 4 π r + 2 π h o r \frac{d S}{d h} = 2 π r

im not sure what to do from here .

I think I need more information .

I^{'} m unsure

- c o n t i n u e -

Answered by prakash jain last updated on 26/Dec/15

Ratio = \frac{V}{S} = \frac{π r^{2} h}{2 π r (r + h)} = \frac{r h}{r + h}

S = 2 π r^{2} + 2 π r h \Rightarrow h = \frac{S - 2 π r^{2}}{2 π r}

Taking S constant find r so that V is maximum

f (r) = \frac{r \frac{S - 2 π r^{2}}{2 π r}}{r + \frac{S - 2 π r^{2}}{2 π r}} = \frac{r S - 2 π r^{2}}{S}

f (r) = \frac{1}{S} (r S - 2 π r^{2})

f^{'} (r) = 1 - \frac{4 π r}{S}, f^{″} (r) = - \frac{4 π}{S}

1 - \frac{4 π r}{S} = 0 \Rightarrow r = \frac{S}{4 π}

Commented by RasheedSindhi last updated on 26/Dec/15

T h a n k s a l o t!

P e r h a p s a c o m e r c i a l p e r s o n

p r e f e r t h e a n s w e r i n d i a m e t e r

a n d h e i g h t r e l a t i o n!

Commented by Filup last updated on 27/Dec/15

r = \frac{S}{4 π}

r = \frac{r^{2}}{2} + \frac{r h}{2}

2 - r = h

Commented by Rasheed Soomro last updated on 27/Dec/15

Thanks!