“微信扫一扫”进入题库练习及模拟考试
发电站建在距
参考答案:\(\mathrm{y}=\mathrm{}\frac {15} {2}{x}^{2}-500x+25000=\frac {15} {2}\left ( {x-\frac {100} {3}} \right )^{2}+\frac {50000} {3}\),当\( \mathrm{x}=\frac{100}{3}\)时,\(y\)取得最小值,\({y}_{\text{min}}=\frac {50000} {3}\),故发电站建在距\(A\)城\( \frac{100}{3}\) km处,能使月供电总费用\(y\)最少.