To find the dual problems for the given primal problems, we will follow the rules for constructing duals from primal linear programming problems.
Primal Problem 1: \[ \text{Maximize } z = 2x_1 + 10x_2 \] Subject to: \[ \begin{align*} 2x_1 + 5x_2 & \leq 16 \quad (1) \ 6x_1 & \leq 30 \quad (2) \ x_1, x_2 & \geq 0 \end{align*} \]
Step 1: Identify the coefficients for dual formulation.
- The objective function has coefficients \(c_1 = 2\) and \(c_2 = 10\).
- The constraints have the right-hand side (RHS) values \(b_1 = 16\) and \(b_2 = 30\).
- The coefficients of the primal constraints form the matrix: \[ \begin{bmatrix} 2 & 5 \ 6 & 0 \end{bmatrix} \]
Dual Problem \[ \text{Minimize } w = 16y_1 + 30y_2 \] Subject to: \[ \begin{align*} 2y_1 + 6y_2 & \geq 2 \quad (1) \ 5y_1 & \geq 10 \quad (2) \ y_1, y_2 & \geq 0 \end{align*} \]
Primal Problem 2: \[ \text{Minimize } z = 5x_1 + 7x_2 \] Subject to: \[ \begin{align*} x_1 + 2x_2 & \leq 20 \quad (1) \ 3x_1 + x_2 & \geq 15 \quad (2) \ 4x_1 + 3x_2 & \leq 60 \quad (3) \ x_1, x_2 & \geq 0 \end{align*} \]
Step 2: Identify the coefficients for dual formulation.
- For the objective function, we have \(c_1 = 5\) and \(c_2 = 7\).
- The constraints have the RHS values \(b_1 = 20\), \(b_2 = 15\), and \(b_3 = 60\).
- The coefficients of the primal constraints form the matrix: \[ \begin{bmatrix} 1 & 2 \ 3 & 1 \ 4 & 3 \end{bmatrix} \]
Dual Problem \[ \text{Maximize } w = 20y_1 + 15y_2 + 60y_3 \] Subject to: \[ \begin{align*} y_1 + 3y_2 + 4y_3 & \leq 5 \quad (1) \ 2y_1 + y_2 + 3y_3 & \leq 7 \quad (2) \ y_1 & \geq 0 \ y_2 & \text{ free} \ y_3 & \geq 0 \end{align*} \]
To summarize, the dual problems for the given primal problems are:
Dual Problem for Primal Problem 1: \[ \text{Minimize } w = 16y_1 + 30y_2 \] Subject to: \[ \begin{align*} 2y_1 + 6y_2 & \geq 2 \ 5y_1 & \geq 10 \ y_1, y_2 & \geq 0 \end{align*} \]
Dual Problem for Primal Problem 2: \[ \text{Maximize } w = 20y_1 + 15y_2 + 60y_3 \] Subject to: \[ \begin{align*} y_1 + 3y_2 + 4y_3 & \leq 5 \ 2y_1 + y_2 + 3y_3 & \leq 7 \ y_1 & \geq 0 \ y_2 & \text{ free} \ y_3 & \geq 0 \end{align*} \]