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標題:
數學題 plz help
發問:
有n個整數,它們的值可能是-1,0,1,2之一.它們之和是19,而平方和為99. 求它們立方和的最大值和最小值.
最佳解答:
假設有 p 個 2, q 個 1, r 個 -1, (幾多個 0 對結果冇影響) 所以 2p + q - r = 19 . . . .(1) 4p + q + r = 99 . . . (2) 8p + q - r = M ==> M = 6p + (2p + q - r) = 6p + 19 所以 M 的最大值是有最多個 2，及最小值是有最小個 2。 從 (1) 得 : q - r = 19 - 2p 從 (2) 得 : q + r = 99 - 4p 所以 q = 59 - 3p 及 r = 40 - p 因為 p, q, r 最小是 0, 所以 p 最小是 0, 最大是 19. 當 p 最大時, 即 p = 19, 這時 q = 59 - 3*19 = 2, r = 40 - 19 = 21, M = 6*17 + 19 = 133 當 p 最小時, 即 p = 0, 這時 q = 59 - 3*0 = 59, r = 40 - 0 = 40, M = 6*0 + 19 = 19 所以它們立方和的最大值是 133，最小值是 19。
其他解答:
Nice question! I have been thinking Cauchy-Schwarz, Holder, Minkowsky, Generalized Mean Ineq. etc. But I am still thinking the way related to the discrete choices of -1, 0, 1, 2. Maybe let others have a try. Quite tired and frustrated, as you can see my carelessness in your next question 2013-09-18 13:42:28 補充： 很好，現在方便一點～
數學題 plz help
發問:
有n個整數,它們的值可能是-1,0,1,2之一.它們之和是19,而平方和為99. 求它們立方和的最大值和最小值.
最佳解答:
假設有 p 個 2, q 個 1, r 個 -1, (幾多個 0 對結果冇影響) 所以 2p + q - r = 19 . . . .(1) 4p + q + r = 99 . . . (2) 8p + q - r = M ==> M = 6p + (2p + q - r) = 6p + 19 所以 M 的最大值是有最多個 2，及最小值是有最小個 2。 從 (1) 得 : q - r = 19 - 2p 從 (2) 得 : q + r = 99 - 4p 所以 q = 59 - 3p 及 r = 40 - p 因為 p, q, r 最小是 0, 所以 p 最小是 0, 最大是 19. 當 p 最大時, 即 p = 19, 這時 q = 59 - 3*19 = 2, r = 40 - 19 = 21, M = 6*17 + 19 = 133 當 p 最小時, 即 p = 0, 這時 q = 59 - 3*0 = 59, r = 40 - 0 = 40, M = 6*0 + 19 = 19 所以它們立方和的最大值是 133，最小值是 19。
其他解答:
Nice question! I have been thinking Cauchy-Schwarz, Holder, Minkowsky, Generalized Mean Ineq. etc. But I am still thinking the way related to the discrete choices of -1, 0, 1, 2. Maybe let others have a try. Quite tired and frustrated, as you can see my carelessness in your next question 2013-09-18 13:42:28 補充： 很好，現在方便一點～
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