由 galaxylee 於 星期日 三月 19, 2006 2:28 pm
#ed_op#DIV#ed_cl##ed_op#DIV#ed_cl#設底半徑a公尺,高b公尺#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#條件:(πa^2)*(b)=(3/2)π => ba^2=3/2#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#要求12*(πa^2)+8*(2aπ)*b=4π(3a^2+4ab)=12π(a^2+2/a)之最小值#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#由算幾不等式#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#a^2+1/a+1/a≧3,等號成立時,a^2=1/a=1/a=1 =>a=1,b=3/2#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#圓柱底半徑1公尺,高3/2公尺,成本36π元為最低#ed_op#/DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#