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2題 F.4 Maths 20分 明天考試用!

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http://i.na.cx/5D5M8.jpg 只係c2 http://i.na.cx/3989b.jpg 只係62 唔該曬!!!明天考試用

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c2. For x pack of A, there are 6x cans of drinks. For y pack of B, there are 8y cans of drinks. So total no. of drinks = 6x + 8y Total cost = 18x + 27y Therefore mean cost = cost per drink = (18x + 27y)/(6x + 8y) For x = y = 4, mean cost = 3.21 For x = 4, y = 5, mean cost = [18(4) + 27(5)]/[6(4) + 8(5)] = 3.23 For x = 5, y = 4, mean cost = [18(5) + 27(4)]/[6(5) + 8(4)] = 3.19 For x = 6, y = 4, mean cost = [18(6) + 27(4)]/[6(6) + 8(4)] = 3.17 For x = y = 5, mean cost = 3.23. So minimum mean cost is at x = 6 and y = 4 Total cost = 18(6) + 27(4) = 216 No. of cans = 6(6) + 8(4) = 68. 2014-06-16 07:22:12 補充： 62. 2/(k - 2) - 1/(k + 1) = 4/5 Multiply each of the 3 terms by 5(k - 2)(k + 1), we get 10(k + 1) - 5(k - 2) = 4(k + 1)(k - 2) 10k + 10 - 5k + 10 = 4k^2 - 8 - 4k 4k^2 - 9k - 28 = 0 (4k + 7)(k - 4) = 0 k = - 7/4 or 4. 2014-06-16 07:24:40 補充： Correction to c2 : the mean cost at x = y = 5 should be the same as the mean cost at x = y = 4, that is 3.21, not 3.23.

其他解答:

↑↑↑↑↑↑↑↑↑↑↑So GOOD!!!!!!!!!!!!!!!!↑↑↑↑↑↑↑↑↑↑↑|||||62) f(x) = 1/(x+1) 2f(k-3)-f(k) = 4/5 2f(k-3)-f(k) = 2[1/(k-3+1)] - 1/(k+1) = 4/5 2[1/(k-3+1)] - 1/(k+1) = 4/5 2[1/(k-2)] - 1/(k+1) = 4/5 2/(k-2) - 1/(k+1) = 4/5 通分母and solve it by yourself
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