2題 F.4 Maths 20分 明天考試用!

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.

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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