(1)据题意,商品的日销售额F(t)=f(t)g(t), 得F(t)= | (t+20)(-t+30),0≤t<10,t∈N | (-t+40)(-t+30),10≤t≤20,t∈N |
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即F(t)= | -t2+10t+600,(0≤t<10,t∈N) | t2-70t+1200,(10≤t≤20,t∈N) |
| | (6分) (2)当0≤t<10,t∈N时, F(t)=-t2+10t+600=-(t-5)2+625, ∴当t=5时,F(t)max=625; 当10≤t≤20,t∈N时, F(t)=t2-70t+1200=(t-35)2-25, ∴当t=10时,F(t)max=600<625 综上所述,当t=5时,日销售额F(t)最大,且最大值为625.(12分) |