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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< k < (sqrt10)/3? 其實我還是很混亂,煩請你續指教我(b)部份吧! (b) Let k=2. Find, in terms of a, the minimum value of T for (i) b = 2a (ii) b = a/2
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, 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.
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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< k < (sqrt10)/3? 其實我還是很混亂,煩請你續指教我(b)部份吧! (b) Let k=2. Find, in terms of a, the minimum value of T for (i) b = 2a (ii) b = a/2
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, 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.
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