(1)设产品每吨价格上涨x%时,销售总金额为y元. 则y=10(1+x%)′1000(1-mx%) =-mx2+100(1-m)x+10000 当m=时,y=-(x-50)2+11250, 故当x=50时,ymax=11250(元). (2)y=-mx2+100(1-m)x+10000x∈(0,80] y=-mx2+100(1-m)x+10000>10000在x∈(0,80]恒成立 除去x得,-mx+100(1-m)>0在x∈(0,80]恒成立 m为正常数,>x, 而x∈(0,80], 故>80, ∴m∈(0,). |