maths plz help

2012-11-26 3:28 am
A,B,C,D是整數,且-2<=A,B,C,D<=8
若A+B+C+D=10 ,求A-2B+3C-4D的最大和最小值.

回答 (2)

2012-11-26 5:25 pm
✔ 最佳答案
沒什麼頭緒0 0....
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A + B + C + D = 10
則 A = 10 - B - C - D ... (i)


P = A - 2B + 3C - 4D
答案 = 找出 P的最大和最小值

代 (i) 入 P:
P
= (10 - B - C - D) - 2B + 3C - 4D
= 10 + 2C - (3B + 5D) ...(ii)

Case 1, P的最大值:
根據(ii):
C越大, B,D 越小, P就越大
所以,
當 C = 8, B = D = -2
則 (i): A = 10 - (-2) - (-2) - 8
-> A = 10 + 2 + 2 - 8 = 6
->
P的最大值 = 6 - 2(-2) + 3(8) - 4(-2) = 42

P = A - 2B + 3C - 4D
= = = = =

Case 2, P的最小值:
A + B + C + D = 10
-> B = 10 - A - C - D ...(iii)

代 (iii) 入 P
P
= A - 2(10 - A - C - D) + 3C - 4D
= (3A + 5C) - 2D - 20 ... (iv)

根據 (iv):
D越大, A、C越小 則P越小
所以, 當D = 8, A = C = -2
則 (ii): B = 10 - (-2) - (-2) - 8 = 6
->
P的最小值 = (-2) - 2(6) + 3(-2) - 4(8) = -52

所以
A - 2B + 3C - 4D 的最大值 = 42 (同時, A = 6, B = -2, C = 8, D = -2)
A - 2B + 3C - 4D 的最小值 = -52 (同時, A = -2, B = 6, C = -2, D = 8)

這樣? 0.0
參考: 自己
2012-11-26 3:22 pm
Assume A, B, C and D are not the same.
A = 6, B = -1, C = 7 and D = - 2
A - 2B + 3C - 4D = 6 - (-1) + 3(7) - 4(-2) = 6 + 1 + 21 + 8 = 36.
A = - 1, B = 6, C = - 2, D = 7
A - 2B + 3C - 4D = - 1 - 2(6) + 3(-2) - 4(7) = - 1 - 12 - 6 - 28 = - 47.

2012-11-26 07:25:25 補充:
Correction : For 1st part, it is even higher if C = 8 and A = 5, answer = 37.

2012-11-26 07:26:58 補充:
Sorry, answer should be 38.


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