 Solution

The two primal constraints are equalities, so let the two dual variables $u,v$ be free. Primal feasibility restricts $x$ to the nonnegative orthant, givingThus the feasible region lies above all four lines:Its lower boundary follows the third line up to $(5/9,7/9)$, the fourth line from there to $(10/3,-1/3)$, and then the first line. The second line never determines the envelope. Along these three pieces $u+v$ respectively has slopes $-3$, $3/5$, and $3/4$, so its unique minimum is at the first corner:

At this dual point the first two inequalities are strict, while the last two are equalities. complementary slackness therefore suggests $x_1=x_2=0$. The remaining primal equations become $4x_3+2x_4=300$, $x_3+5x_4=300$, giving $x_3=x_4=50$. This is nonnegative and satisfies both equations. Its objective function is $3\cdot 50+5\cdot 50=400$, equal to the feasible dual value. weak duality provesNo nonnegativity restriction was imposed on the dual variables; the negative-$v$ second corner is part of the feasible boundary, even though it is not optimal.