(1)解:∵甲地一共需要27台,已经从A省运来x台,
∴从B应运来(27-x)台,
∵乙地需要25台,已经从B地运来(x-x)台,
∴应从A地再运来25-(x-p)=28-x(台),
故答案为:27-x,28-x;
(2)解:根据题意得:3.4x+3.5(27-x)+3.3(28-x)+3.2(x-3)≤16.2,
x≥25.5,
∵27-x≥0,28-x≥0,x-3≥0,
∴x的取值范围是25.5≤x≤27,
∵x表示机器的台数,
∴x只能取26和27,
即有两种调运方案;
①当x=26时,27-x=1,28-x=2,x-0=20,
即从A省运往甲地26台,运往乙地2台,而从B省运往甲地1台,运往乙地23台,
y=0.4×26+0.5×1+0.3×2+0.2×23=16.1(万元);
②当x=27时,27-x=0,28-x=1,x-3=24,
即从A省运往甲地27台,运往乙地p台,而从B省运往甲地0台,运往乙地24台,
y=0.4×27+0.5×0+0.3×h+0.2×24=h5.9(万元)<h6.h(万元),
即若要使总耗资不超过16.2万元,调运方案是①从A省运往甲地26台,运往乙地2台,而从B省运往甲地1台,运往乙地23台,②从A省运往甲地27台,运往乙地1台,而从B省运往甲地0台,运往乙地24台,①种调运方案的总耗资最少.
你这是数学题,应请教专业人士