# Standard form and basic solutions

Is it possible that for a general linear program none of the optimal solutions have to be basic? Meanwhile, for a program in standard form an optimal solution has be basic. If so, why ?

For $$\min_{x \geq 0} \{ x : x \leq 1\}$$ the only basic feasible solution is $x=1$, which is not optimal.
• Why $x = 0$ is not a basic feasible solution? If I add a slack variable $s \ge 0$, then $x + s = 1$. The optimal solution $(x,s) = (0,1)$ has one non-zero entry, so it's basic. – GNUSupporter 8964民主女神 地下教會 Apr 6 '18 at 17:18
• @GNUSupporter That's a different polygon. I am looking at $Ax \leq b$, not $Ax + s = b$ (even though its projection on $x$ is the same). Also see this definition of basic feasible solution. – LinAlg Apr 6 '18 at 17:42