Question Number 141368 by bemath last updated on 18/May/21
$${A}\:{closed}\:{cylindrical}\:{can}\:{be}\:{is}\:{to}\:{hold} \\ $$$$\mathrm{1}\:{liters}\:{of}\:{liquid}\:.\:{How}\:{should}\:{we}\: \\ $$$${choose}\:{the}\:{height}\:{and}\:{radius}\: \\ $$$${to}\:{minimize}\:{the}\:{amount}\:{of} \\ $$$${material}\:{needed}\:{to}\:{manufacture} \\ $$$${the}\:{can}\:?\: \\ $$
Answered by TheSupreme last updated on 18/May/21
$${you}\:{have}\:{to}\:{minimize} \\ $$$${S}=\mathrm{2}\pi{R}^{\mathrm{2}} +\pi{RH} \\ $$$${under}\:{the}\:{constrain} \\ $$$${V}=\pi{R}^{\mathrm{2}} {H}=\mathrm{1}{lt} \\ $$$${H}=\frac{{V}}{\pi{R}^{\mathrm{2}} } \\ $$$${S}=\mathrm{2}\pi{R}^{\mathrm{2}} +\frac{{V}}{{R}} \\ $$$${min}\left({S}\right) \\ $$$$\frac{\partial{S}}{\partial{R}}=\mathrm{4}\pi{R}−\frac{{V}}{{R}^{\mathrm{2}} }=\mathrm{0}\:\rightarrow\:{R}=^{\mathrm{3}} \sqrt{\frac{{V}}{\mathrm{4}\pi}} \\ $$$${R}=\mathrm{4}.\mathrm{30}\:{cm} \\ $$$${H}=\mathrm{17}.\mathrm{21}\:{cm} \\ $$$$ \\ $$
Commented by iloveisrael last updated on 18/May/21
$${r}\:\approx\:\mathrm{5}.\mathrm{4}\:{cm} \\ $$
Answered by iloveisrael last updated on 18/May/21
