问题 解答题

某电视台与某广告公司约定播放甲、乙两部电视剧,经调查,播放甲连续剧平均每集有观众20万人次,播放乙连续剧平均每集有观众15万人次,公司要求电视台每周共播放7集。

(1)设一周内甲连续剧播x集,甲、乙两部连续剧的观众总收视人数为y万人次,求y与x的函数关系式;

(2)已知电视台每周只能为该公司提供不超过300分钟的播放时间,并且播放甲连续剧每集50分钟,播放乙连续剧每集35分钟,问电视台每周应各播放甲、乙两种连续剧多少集,才能使每周收视观众的人数总和最大?并求出这个最大值。

答案

(1)y=5x+105;(2)甲3集,乙4集,最大值是120万人次

题目分析:(1)根据等量关系:一周内观看甲连续剧的人数+观看乙连续剧的人数=总人数,即可列出函数关系式;

(2)根据甲连续剧播放时间+乙连续剧时间≤300分钟,即可得到x的范围,再结合(1)中的函数关系式的性质即可得到结果.

(1)由题意得y=20x+15(7-x)

∴y=5x+105;

(2)由题意得50x+35(7-x)≤300

解得

又y=5x+105的函数值随着x的增大而增大.

又∵x为自然数,

当x=3时,y有最大值3×5+105=120(万人次)

7-x=4

答:电视台每周应播出甲连续剧3集,播放乙连续剧4集,才能使每周收视观众的人次总和最大,这个最大值是120万人次.

点评:函数的应用是初中数学的重点和难点,是中考常见题,在压轴题中极为常见,难度较大.

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