Question

高一函数题！急！！！！

某种商品在近100天内（包括100天）的销售量和价格（元/件）都是时间t的函数，且销售量近似的满足关系式g(t)=-1/3t+109/3。在前40天（包括第40天），每天的价格近似的满足函数f(t)=1/4t+22;在后60天，每天的价格近似的满足函数f(t)=-1/2t+52，求这种商品日销售额的最大值（近似到元） ~~
速度....
详细点..要结果的!

Answer 1

t g(t) f(t) 当日销售额个g(t)*f(t)

1 36.0000 22.2500 801.0000
2 36.1667 22.1250 800.1875
3 36.2222 22.0833 799.9074
4 36.2500 22.0625 799.7656
5 36.2667 22.0500 799.6800
6 36.2778 22.0417 799.6227
7 36.2857 22.0357 799.5816
8 36.2917 22.0313 799.5508
9 36.2963 22.0278 799.5267
10 36.3000 22.0250 799.5075
11 36.3030 22.0227 799.4917
12 36.3056 22.0208 799.4786
13 36.3077 22.0192 799.4675
14 36.3095 22.0179 799.4579
15 36.3111 22.0167 799.4496
16 36.3125 22.0156 799.4424
17 36.3137 22.0147 799.4360
18 36.3148 22.0139 799.4303
19 36.3158 22.0132 799.4252
20 36.3167 22.0125 799.4206
21 36.3175 22.0119 799.4165
22 36.3182 22.0114 799.4127
23 36.3188 22.0109 799.4093
24 36.3194 22.0104 799.4061
25 36.3200 22.0100 799.4032
26 36.3205 22.0096 799.4005
27 36.3210 22.0093 799.3980
28 36.3214 22.0089 799.3957
29 36.3218 22.0086 799.3936
30 36.3222 22.0083 799.3916
31 36.3226 22.0081 799.3897
32 36.3229 22.0078 799.3879
33 36.3232 22.0076 799.3863
34 36.3235 22.0074 799.3847
35 36.3238 22.0071 799.3833
36 36.3241 22.0069 799.3819
37 36.3243 22.0068 799.3806
38 36.3246 22.0066 799.3793
39 36.3248 22.0064 799.3782
40 36.3250 22.0063 799.3770

41 36.3252 51.9878 1888.4676
42 36.3254 51.9881 1888.4882
43 36.3256 51.9884 1888.5078
44 36.3258 51.9886 1888.5266
45 36.3259 51.9889 1888.5445
46 36.3261 51.9891 1888.5617
47 36.3262 51.9894 1888.5781
48 36.3264 51.9896 1888.5938
49 36.3265 51.9898 1888.6089
50 36.3267 51.9900 1888.6234
51 36.3268 51.9902 1888.6373
52 36.3269 51.9904 1888.6507
53 36.3270 51.9906 1888.6636
54 36.3272 51.9907 1888.6760
55 36.3273 51.9909 1888.6879
56 36.3274 51.9911 1888.6995
57 36.3275 51.9912 1888.7106
58 36.3276 51.9914 1888.7213
59 36.3277 51.9915 1888.7317
60 36.3278 51.9917 1888.7417
61 36.3279 51.9918 1888.7514
62 36.3280 51.9919 1888.7608
63 36.3280 51.9921 1888.7699
64 36.3281 51.9922 1888.7787
65 36.3282 51.9923 1888.7872
66 36.3283 51.9924 1888.7955
67 36.3284 51.9925 1888.8035
68 36.3284 51.9926 1888.8113
69 36.3285 51.9928 1888.8189
70 36.3286 51.9929 1888.8262
71 36.3286 51.9930 1888.8334
72 36.3287 51.9931 1888.8403
73 36.3288 51.9932 1888.8471
74 36.3288 51.9932 1888.8536
75 36.3289 51.9933 1888.8600
76 36.3289 51.9934 1888.8663
77 36.3290 51.9935 1888.8723
78 36.3291 51.9936 1888.8782
79 36.3291 51.9937 1888.8840
80 36.3292 51.9938 1888.8896
81 36.3292 51.9938 1888.8951
82 36.3293 51.9939 1888.9004
83 36.3293 51.9940 1888.9056
84 36.3294 51.9940 1888.9107
85 36.3294 51.9941 1888.9157
86 36.3295 51.9942 1888.9206
87 36.3295 51.9943 1888.9253
88 36.3295 51.9943 1888.9299
89 36.3296 51.9944 1888.9345
90 36.3296 51.9944 1888.9389
91 36.3297 51.9945 1888.9432
92 36.3297 51.9946 1888.9475
93 36.3297 51.9946 1888.9516
94 36.3298 51.9947 1888.9557
95 36.3298 51.9947 1888.9597
96 36.3299 51.9948 1888.9636
97 36.3299 51.9948 1888.9674
98 36.3299 51.9949 1888.9711
99 36.3300 51.9949 1888.9748
100 36.3300 51.9950 1888.9784 *******

两段函数在各自给定的（天数）定义域内均为单调函数
取值：1889元/天

Answer 2

销售量满足关系式g(t)=-1/3t+109/3
在前40天(包括第40天),每天的价格近似的满足函数f1(t)=1/4t+22 (0≤t≤40)
在后60天，每天的价格近似的满足函数f2(t)=-1/2t+52 (40≤t≤100)
要求这种商品日销售额的最大值
将g(t)=-1/3t+109/3与f1(t)=1/4t+22相乘
然后配成顶点式即可知其最值 (顶点式即y=a(x-xo)^2+h)
一般式y=ax^2+bx+c
顶点式的xo=-b/2a h=(4ac-b^2)/4a
注意范围还是(0≤t≤40)
同样将g(t)=-1/3t+109/3与f2(t)=-1/2t+52
注意范围还是(40≤t≤100)
两者比较即可得这种商品日销售额的最大值

大概告诉你个方法好啦,自己算算吧,不能太依赖了，连结果自己都懒算了这样不好哦

Answer 3

日销售额=销售量×价格=g(t)×f（t）
只是价格关于时间这个函数是 分段函数，不知道你学过分段函数这个概念没。简而言之就是要把时间分成两个部分来算，也就是前40天的销售额=g（t）×（1/4t+22),后六十天的=g(t)×（-1/2t+52)
然后计算就ok了
一元二次函数的最值应该学过吧，好像前面最大值在t=1处=799 后面最大值是359，你再自己算一次吧，

Answer 4

哇……

不会……