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A.Maths - Rates of change
發問:
http://good-times.webshots.com/photo/2428570460103642988OASDyd (b) Let k=2. Find, in terms of a, the minimum value of T for (i) b = 2a (ii) b = a/2 只想請教(a)(iii) and (b) 更新: 題目不是早說了k>1麼?那麼 range of k 是否該 1
最佳解答:
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, 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 3a, range of k is -(sqrt10)/3 0, range of k becomes 0
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