The Dual of the given Primal is : Maximise : Z = 40x_1 + 35x_2 S.t. : 6x_1 + 4x_2 ≤ 220 3x_1 + 10x_2 ≤…

For a standard primal maximization problem $\max Z = c^T x$ subject to $Ax \le b$ and $x \ge 0$, the dual is the minimization problem $\min C = b^T y$ subject to $A^T y \ge c$ and $y \ge 0$. Interchanging the objective coefficients $(40, 35)$ with the right-hand-side constants $(220, 280)$, transposing the constraint matrix, and reversing the inequalities yields "Minimum : C = 220y_1 + 280y_2, S.t. : 6y_1 + 3y_2 ≥ 40; 4y_1 + 10y_2 ≥ 35; y_1, y_2 ≥ 0".