由题意知:存款量f(x)=kx2,当利率为0.012时,存款量为1.44亿,由1.44=k•(0.012)2,得k=10000,得f(x)=10000x2,银行应支付的利息g(x)=x•f(x)=10000x3,设银行可获收益为y,则y=480x2-10000x3,再由导数性质能求出当x为多少时,银行可获得最大收益.
【解析】
由题意知:存款量f(x)=kx2,
当利率为0.012时,存款量为1.44亿,
即x=0.012时,y=1.44;
由1.44=k•(0.012)2,得k=10000,
∴f(x)=10000x2,
银行应支付的利息g(x)=x•f(x)=10000x3,
设银行可获收益为y=贷款收益-利息支出,
则y=480x2-10000x3,
由于y'=960x-30000x2,则y'=0,
即960x-30000x2=0,得x=0或x=0.032.
因为x∈(0,0.032)时,y'>0,
此时,函数y=480x2-10000x3递增;
x∈(0.032,0.048)时,y'<0,
此时,函数y=480x2-10000x3递减;
故当x=0.032时,y有最大值,其值约为0.164亿.