A closed cylindrical can be is to hold 1 liters of liquid . How should we choose the height and radius to minimize the amount of material needed to manufacture the can ?


Question and Answers Forum

All Questions Topic List
Differentiation Questions
Previous in All Question Next in All Question
Previous in Differentiation Next in Differentiation
Question Number 141368 by bemath last updated on 18/May/21

$A c l o s e d c y l i n d r i c a l c a n b e i s t o h o l d$ $1 l i t e r s o f l i q u i d . H o w s h o u l d w e$ $c h o o s e t h e h e i g h t a n d r a d i u s$ $t o m i n i m i z e t h e a m o u n t o f$ $m a t e r i a l n e e d e d t o m a n u f a c t u r e$ $t h e c a n ?$
	Answered by TheSupreme last updated on 18/May/21

	$y o u h a v e t o m i n i m i z e$ $S = 2 π R^{2} + π R H$ $u n d e r t h e c o n s t r a i n$ $V = π R^{2} H = 1 l t$ $H = \frac{V}{π R^{2}}$ $S = 2 π R^{2} + \frac{V}{R}$ $m i n (S)$ $\frac{\partial S}{\partial R} = 4 π R - \frac{V}{R^{2}} = 0 \to R =^{3} \sqrt{\frac{V}{4 π}}$ $R = 4.30 c m$ $H = 17.21 c m$
	Commented by iloveisrael last updated on 18/May/21

	$r \approx 5.4 c m$
	Answered by iloveisrael last updated on 18/May/21

Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum