初三 数学应用题

如题所述

1、(1)涨价:y=300-10(x-50)

y=800-10x  (50<x<80)

(2)降价:y=300+20(50-x)

y=1300-20x(40<x<50)

2、(1)涨价:

W=(X-40)[300-10(x-50)]

W=-10x²+1200x-32000

(2)降价:

W=(50-x)[300+20(50-x)]

W=20x²-2300x+65000

3、(1)涨价:

W=-10x²+1200x-32000

   =-10(X²-120X+3600)+4000

=-10(X-60)²+4000

X=60时,有最大值4000

∵x≤56   (40+40×40%=56元)

∴x=56的利润:

W=-10(56-60)²+4000

    =3840元

(2)降价:

W=20x²-2300x+65000

   =20(x²+115x+13335/4)-66125+65000

   =20(x+115/4)²-1125

无最大值,

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