高難度二元一次不等式

首先能得出關係式

25x + 40y >=300
x + y >=12
x>=2
y>=2

畫圖後(欠圖請見諒),得到幾點符合條件的坐標.
(2,7)(2,8)(2,9)(2,10)(3,6)(3,7)(3,8)(3,9)(4,5)(4,6)(4,7)(4,8)(5,5)(5,6)(5,7)(6,4)(6,5)(6,6)(7,3)(7,4)(7,5)(8,2)(8,3)(8,4)(9,2)(9,3)(10,2)

消費
(2,7)=352

(2,8)=392

(2,9)=432

(2,10)=472

(3,6)=348

(3,7)=388

(3,8)=428

(3,9)=468

(4,5)=344

(4,6)=384

(4,7)=424

(4,8)=464

(5,5)=380

(5,6)=420

(5,7)=460

(6,4)=376

(6,5)=416

(6,6)=456

(7,3)=372

(7,4)=412

(7,5)=452

(8,2)=368

(8,3)=408

(8,4)=448

(9,2)=404

(9,3)=444

(10,2)=440

由此可見,當x=4,y=5時
可得最少消費$344.

25x+40y>=300..........................x+1.6y>=300........................x>=300-1.6y
36x+25y<=360..........................6^2 x+5^2 y<=360
x>=2
y>=2


36x+25y<=360
x=4
y=8

呢個唔係二元一次不等式,應該係線性規劃

約束條件
設x 為A 糖果,y為B糖果
x≧2
y≧2
x,y 為整數
25x + 40y≧300
36x+40y≦60
x+y≧12

幅圖我唔sd上黎喇,因為好難畫..sorry!!

2008-12-24 20:38:53 補充：
25x +40y 大過等於300
36x+40y細過等於360
x+y細過等於12

先弄個model出來
設小明買x 包糖果A y包糖果B
則MODEL是
MIN 36x+40y
s.t.
25x+40y>=300 (糖果數)
36x+40y<=360 (預算)
x>=2, y>=2 (條件)
x+y<=12 (條件)
x,y整數
最後得x=2,y=7
最低消費352元