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歡迎您訪問數學考題練習:某商店經銷一種成本為每千克20元的水產品,據市!

數學考題練習:某商店經銷一種成本為每千克20元的水產品,據市

更新時間:2024-01-12 14:09:23作者:貝語網校

某商店經銷一種成本為每千克20元的水產品,據市場分析,若按每千克30元銷售,一個月能售出500kg,銷售單價每漲1元,月銷售量就減少10kg,解答以下問題.

(1)當銷售單價定位每千克35元時,計算銷售量和月銷售利潤;

(2)設銷售單價為x元,月銷售收入為y元,請求出y與x的函數關系;

(3)商店想在月銷售成本不超過6000元的情況下,使得月銷售利潤達到8000元,銷售單價應為多少?

試題答案

解:(1)銷售量:500-5×10=450(kg);

銷售利潤:450×(35-20)=450×15=6750元

(2)y=(x-20)[500-10(x-30)]=-10x2+1000x-16000

(3)由于水產品不超過10000÷40=250kg,定價為x元,

則(x-20)[500-10(x-30)]=8000

解得:x1=40,x2=60

當x1=40時,進貨500-10(40-30)=400kg>300kg,舍去,

當x2=60時,進貨500-10(60-30)=200kg<300kg,符合題意.

試題解析

(1)根據題意計算即可;

(2)利潤=銷售量×單位利潤.單位利潤為x-20,銷售量為500-10(x-30),據此表示利潤得關系式;

(3)銷售成本不超過6000元,即進貨不超過6000÷20=300kg.根據利潤表達式求出當利潤是8000時的售價,從而計算銷售量,與進貨量比較得結論.

點評:本題考查了一元二次方程的應用,此題的創意在第三問,同時考慮進出兩個方面的問題,比較后得結論.

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