Question 672605
an aluminum can us being designed to hold 17013pi cubic centimeters of water.
 Find the dimensions of the can that result in the lest amount of aluminum used
 in the construction of the can
:
Volume
{{{pi*r^2*h}}} = {{{17013*pi}}}
divide both sides by pi
{{{r^2*h}}} = 17013
divide both sides by r^2
h = {{{17013/r^2}}}

:
Surface area of a cylinder; (including the ends)
S.A. = {{{2*pi*r(r + h)}}}
:
Replace h with  {{{17013/r^2}}}
S.A = {{{2*pi*r(r+(17013/r^2))}}} =  {{{2*pi*(r^2+r(17013/r^2))}}} =  {{{2*pi*(r^2+(17013/r))}}}
Graphically
{{{ graph( 300, 200, -10, 50, -5000, 15000, 2*pi*(x^2+(17013/x)) ) }}}
Can radius for minimum surface area = 20.41 inches
:
Find the height of the cann
h = {{{17013/20.41^2}}}
h = 40.84 cm high
:
:
:
Check by finding the vol of a can with a radius of 20.41 and a height of 40.84
Vol = {{{pi*20.41^2*40.84}}}
V = 53447 cu/in; (which is {{{17013*pi)}}})