Solution
= Solution
Changing only the second profit coefficient leaves the dual prices at $(1,1)$, because the basic products remain one and three. The nonbasic product has nonpositive <reduced cost> exactly when $c_2\le3(1)+2(1)=5$. Therefore
$$
\boxed{c_2\le5.}
$$
There is no mathematical lower bound; if profit coefficients are additionally restricted to be nonnegative, intersect this range with $[0,\infty)$. At $c_2=5$, alternate optimal solutions are possible.