 Solution

Let $x^{\ast }=(0,3,4,0,5{)}^T$. By strong duality, an optimal dual vector $y^{\ast }$ exists. Complementary slackness at $x^{\ast }$ makes the dual constraints equalities at indices $2,3,5$, where $x_j^{\ast }>0$. The vector $\hat{x}=(0,1,2,0,3{)}^T$ has exactly that same support. It is nonnegative and satisfies $A\hat{x}=\hat{b}$, while $y^{\ast }$ remains dual feasible because dual feasibility depends on $A,c$, not on the right-hand side. HenceThe objective gap is zero, so $\hat{x}$ is optimal for the modified programme. This is support-preserving right-hand-side changes in linear programming.