✔ 最佳答案
Distance travelled by road = AP = sqrt(a^2 + x^2).
Cost for this part of the journey = ksqrt(a^2 + x^2).
Distance travelled by railway = PB = b - x. since cost is $1/km, so
cost for this part of the journey = b - x.
So toal cost, T = ksqrt(a^2 + x^2) + b - x.
dT/dx = kx/sqrt(a^2 + x^2) - 1
Put it to zero, we get
kx = sqrt(a^2 + x^2)
k^2x^2 = a^2 + x^2
(k^2 -1)x^2 = a^2
x = a/sqrt(k^2 - 1), that means T is a minimum when x is at this value.
If x = b = 3a,T is a minimum if travel by road directly from A to B, that means it makes no difference to travel from A to B direct or via P.
so 3a = a/sqrt(k^2 - 1)
3sqrt(k^2 - 1) = 1
9(k^2 -1) = 1
9k^2 = 10
so k = (sqrt10)/3 [negative value of k is rejected because k> 0]
If x < 3a, so T is a minimum if travel via P.
That is a/sqrt(k^2 - 1) < 3a
If x > 3a, T is certainly a minimum if travel directly from A to B, travel via P is certainly more expensive.
That is a/sqrt(k^2 - 1) > 3a
1> 9(k^2 - 1)
9k^2 - 10 < 0
[k + (sqrt10)/3][k - (sqrt10)/3] < 0
so 0< k < (sqrt10)/3.
2008-12-08 11:43:00 補充：
If simply a/sqrt(k^2 - 1) > 3a, range of k is -(sqrt10)/3 < k < (sqrt10)/3. When taking into account that k > 0, range of k becomes 0 < k<(sqrt10)/3.
2008-12-10 16:59:09 補充：
1. Sorry, I overlooked this point, so 1< k < (sqrt10)/3.
For b(i), put k = 2, b = 2a and x = a/sqrt3 into the T expression in line 5, you will get the answer.
2008-12-10 17:02:18 補充：
For b(ii), basically the same way can be done for b = a/2. However, a/2 is shorter than the optimum distance x = a/sqrt3, so I am doubtful if the answer is correct.
