Linear Programming problem

a) Suppose x tonnes and y tonnes of medicine are transported from East House to Healthy Store and Heal Store respectively.
So, (100-x) tonnes and (50-y) tonnes of medicine are transported from West House to Healthy Store and Heal Store respectively.0 <= x <= 100 ............... (1)0 <= y <= 50 ................. (2)For West House, (100-x) + (50-y) <= 40ie. x+y >= 110 ............... (3)For East House, x+y <= 140 .................... (4)
[Transportation fee, for Healthy Store,$60x + $30(100-x)= $(3000 + 30x)For Heal Store,$80y + $20(50-y)= $(1000 + 60y)]

2015-01-25 09:57:18 補充：
Intersection points of (x, y) are (100, 10), (100, 40), (60, 50) (90, 50).
Minimum transportation fee occurs when (x, y) is (100, 10). ie.
$(3000 + 30*100 + 1000 + 60*10)
= $7600
Maximum transportation fee is :
$(3000 + 30*90 + 1000 + 60*50)
= $9700

2015-01-25 09:59:11 補充：
I think part (b) is to calculate the minimum cost of transporting,
so the answer is $7600.

似乎題目應該還有其他資料要提供？