A proper meaningful insight can be gained by graphical representation of the problem:
With[{t = 1.9333},
Show[RegionPlot[
2 x + y + 2 < t && -x - 2 y + 3 < t && -3 x + y < t &&
2 x - 3 y < t, {x, -4, 4}, {y, -4, 4}],
Plot[t - 2 x - 2, {x, -4, 4}, PlotStyle -> Orange],
Plot[ (-t - x + 3)/2, {x, -4, 4}, PlotStyle -> Blue],
Plot[ t + 3 x, {x, -4, 4}, PlotStyle -> Gray],
Plot[ (-t + 2 x)/3, {x, -4, 4}, PlotStyle -> Pink]]]

This is not so satisfying. One constraint is obsolete and may be removed without affecting the solution.
So the problem is equivalent to determine the crossing of the three other boundaries given. Then the enclosed area of the three lines bounded area is zero.
The problem posed here fulfills the conditions named in the Mathematica documentation tutorial Linear Programming. It can be solved therefore both with FindMinimum
and LinearProgramming
.
Since the pink-colored boundary is irrelevant the problem reduces to
FindMinimum[{t,
2 x + y + 2 < t && -x - 2 y + 3 < t && -3 x + y < t}, {t, x, y},
Method -> "LinearProgramming"]
this can be formulated equivalent in
LinearProgramming[{1, 0, 0}, {{1, -2, -1}, {1, 1, 2}, {1, 3, -1}}, {2,
3, 0}, None]
The graphic representation saves time and effort.
(* Out: {29/15, -(2/5), 11/15} *)
If the t-value is altered in the RegionPlot a real region appears for t>29/15 and an error is posed for smaller values. The solution is a point in the case of the three value solution. It is a region for bigger values. So this give insight into the solution of the linear programming problem. With this point, a region emerges in which real solutions exist. This is the smallest set of solutions all other included infinite many points, a region.
That is the solution to the linear programming problem here, not the point alone.