數學變分問題(20點)

1.
(a)
Y 是兩部份之和，一部份隨 X + 1 反變,另一部份隨 Z 正變且隨 X 二次反變。
故此， Y = [k1/(X + 1)] + [k2Z/X²]
其中 k1 及 k2 為常數。

當 X = 1 及 Z = 2 時，Y = 1/2 :
1/2 = [k1/(1 + 1)] + [k22/1²]
(k1/2) + 2k2 = 1/2
k1 + 4k2 = 1 ...... [1]

當 X = 2 及 Z = 3 時，Y = 5/8
5/8 = [k1/(2 + 1)] + [k23/2²]
(k1/3) + (3k2/4) = 5/8
24 * [(k1/3) + (3k2/4)] = 24 * (5/8)
8k1 + 18k2 = 15 ...... [2]

[1] * 8 :
8k1 + 32k2 = 8 ...... [3]

[3] - [2] :
14k2 = -7
k2 = -1/2

把 k2 = -1/2 代入 [1] 中：
k1 + 4(-1/2) = 1
k1 - 2 = 1
k1 = 3

故此： Y = [3/(X + 1)] - (Z/2X²)
Z/2X² = [3/(X + 1)] - Y
Z/2X² = [3/(X + 1)] - [Y(X + 1)/(X + 1)]
Z/2X² = (3 - XY - Y) / (X + 1)
Z = 2X²(3 - XY - Y) / (X + 1)


(b)
當 X = 3，Y = 4 :
Z = 2*3²*[3 - 3*4 - 4] / (3 + 1)
Z = -58.5