(Ⅱ)由(Ⅰ)知①当0≤t≤10时y=-t 2 +10t+1200=-(t-5) 2 +1225函数图象开口向下,对称轴为t=5,该函数在t∈[0,5]递增,在t∈(5,10]递减∴y max =1225(当t=5时取得),y min =1200(当t=0或10时取得)②当10<t≤20时y=t 2 -90t+2000=(t-45) 2 -25图象开口向上,对称轴为t=45,该函数在在t∈(10,20]递减,t=10时,y=1200,y min =600(当t=20时取得)由①②知y max =1225(当t=5时取得),y min =600(当t=20时取得)