# Having difficulty with the Minimize command

I am having trouble getting an answer to this question...Am I missing something that anybody can see.

Simplify[Minimize[{(
x (-a + x))/(-a +
H) + (1 - (-a + x)/(-a + H)) ((y (-b + y))/(-b + H) +
H (1 - (-b + y)/(-b + H))), 0 <= a < 100, 0 <= b < 100,
a < x < 100, b < y <= 100}, {x, y}]]


My problem is: The computer (a small netbook) takes forever and I keep having to abort the evaluation. So I would like to know if it is possible to get an answer maybe on a faster machine or am I doing something wrong

• What is H? Any definitions? And you have wrong syntax - it should be && between the inequalities, not comas. Please check documentation for basic Minimize examples. – Vitaliy Kaurov Jun 16 '12 at 5:44
• @Vitaliy, actually the commas are fine; try Minimize[{x^2, 1 <= x <= 2}, x] for instance. – J. M. will be back soon Jun 16 '12 at 14:10
• @J.M. You are showing coma between optimized function and first constraint - which is fine. I was talking about comas between constraints. So comas do work there - it is not documented (at least to my knowledge). Also replacing logical operators with comas will not always work - like in case of ||. This is syntax Documentation shows: wolfram.com/xid/0bn5wuy-wtuty – Vitaliy Kaurov Jun 16 '12 at 15:24
• @Vitaliy: Yes, the use of commas is an implicit use of &&, which some people might not realize at once. However, last I checked, commas as separators for constraints (again, assuming conjunction!) works nicely; e.g. Minimize[{x^2 + y^2, 2 <= x <= 3, 2 <= y <= 3}, {x, y}] and Minimize[{x^2 + y^2, 2 <= x <= 3 && 2 <= y <= 3}, {x, y}] do the same thing. – J. M. will be back soon Jun 16 '12 at 15:31
• Nothing can really be done here without a definition for H. Judging by the allowed ranges and the general form of the expression, I think H=100 is a reasonable guess. With H=100, the Minimize executes reasonably quickly, but gives a sloppy output (with redundant conditions) amounting to {x -> 37.5+.5a+.125b, y -> 50+.5b} if a < 75+.25b. The minimum is then (16a^2+8a(b+300)+(b-1300)(b+300))/(64(a-100)). – Eric Thewalt Mar 13 '13 at 7:10