(1)设购买甲种树苗$x$株,乙种树苗$y$株,
则列方程组$\left\{\begin{array}{l} x+y=800,\\ 24x+30y=\number{21000},\end{array}\right.$
解得$\left\{\begin{array}{l} x=500,\\ y=300.\end{array}\right.$
答:购买甲种树苗$500$株,乙种树苗$300$株.
(2)设购买甲种树苗$z$株,乙种树苗$(800-z )$株.
则列不等式$85\% z+90\% (800-z)\geqslant 88\% \times 800$.
解得$z\leqslant 320$.
答:甲种树苗至多购买$320$株.
(3)设甲种树苗购买$m$株,使购买树苗的费用为$\number{22080}$元,
则$24m+30(800-m)=\number{22080}$.
解得$m=320$.
$800-320=480$,符合(2)的要求.
答:购甲种树苗$320$株,乙种树苗$480$株时,总费用为$\number{22080}$元.
标签:树苗,购买,number