# Minimize is returning unevaluated for a simple positive integer domain problem [closed]

I don't understand why the following Minimize code is returning unevaluated. I believe I have it coded properly and the expected minimum is $$x=14$$ and $$y=1$$.

 Minimize[{1 + x == 15 y, x >= 14 y}, {x, y}, PositiveIntegers]
(* Minimize[{1 + x == 15 y, x >= 14 y}, {x, y}, PositiveIntegers] *)


Can Minimize not solve this problem?

• What does this give you: Minimize[{1 + x = 15 y, x >= 14 y}, {x, y}, PositiveIntegers] ??
– Syed
Jun 2 at 10:59
• @Syed That syntax returns a "Tag 1+x is protected" message but does return the minimum. The double equal sign however is the standard syntax for an equation and I would prefer adhering to strict syntax and not encountering unnecessary messages or errors with my code. Thanks though.
– josh
Jun 2 at 11:44
• What are you trying to minimize? You only supply two conditions, but no function that shall be minimized. Solve[{1 + x == 15 y, x >= 14 y}, {x, y}, PositiveIntegers] works fine. Jun 2 at 11:50
• @Roman: In the case above, there are other values which satisfy the relations such as $(29,2)$, $(46,3)$ and so on. I am looking for the smallest. This concerns a larger problem of $x+a==ny$ with a set of inequalities $x+a_j>=iy$ with $a,b,a_j,n$ all positive integers. Above is only a simple example and I have some concern or skepticism Solve would not always give me the smallest.
– josh
Jun 2 at 12:06
• The "smallest" by what metric? For example, is (29, 2) "smaller than", "bigger than", or "equal to" (28,3) under the ranking you have in mind? Jun 2 at 20:15

You need to specify what you mean by "smallest solution". If it means that you want the smallest value of $$x+y$$, then you can do
Minimize[{x + y, 1 + x == 15 y && x >= 14 y}, {x, y}, PositiveIntegers]