Question: User: Question 3, 6.3.11
Part 1 of 2
In the following problem, a minimization problem, the corresponding dual problem, and the
Minimize C = 21x1 + 50x2 Maximize P = 36y1 + 51y2
subject to 2x1 + 5x2 \geq 36 subject to 2y1 + 3y2 \leq 21
3x1 + 7x2 \geq 51 5y1 + 7y2 \leq 50
x1, x2 \geq 0 y1, y2 \geq 0
y1 y2 x1 x2 P
0 1 5 -2 0 5
1 0 -7 3 0 3
0 0 3 6 1 363
a. Find the optimal solution of the dual problem
Maximum of P = 363
y1 = 3
y2 =
1. **State the problem:**
We are given a dual linear programming problem:
Maximize $$P = 36y_1 + 51y_2$$
subject to
$$2y_1 + 3y_2 \leq 21$$
$$5y_1 + 7y_2 \leq 50$$
$$y_1, y_2 \geq 0$$
We want to find the optimal solution $(y_1, y_2)$ that maximizes $$P$$.
2. **Given tableau:**
\begin{align*}
&\begin{array}{c|ccccc|c}
& y_1 & y_2 & x_1 & x_2 & P & \text{RHS} \\
\hline
\text{Row 1} & 0 & 1 & 5 & -2 & 0 & 5 \\
\text{Row 2} & 1 & 0 & -7 & 3 & 0 & 3 \\
\text{Row 3} & 0 & 0 & 3 & 6 & 1 & 363 \\
\end{array}
\end{align*}
3. **Interpretation:**
- The tableau shows the current basic variables and their values.
- The last row shows the value of $$P = 363$$.
- From the tableau, the basic variables are $$y_1$$ and $$y_2$$ with values given by the RHS of rows 2 and 1 respectively.
4. **Read values from tableau:**
- From Row 2: $$y_1 = 3$$
- From Row 1: $$y_2 = 5$$
5. **Check feasibility:**
- Check constraints:
- $$2y_1 + 3y_2 = 2(3) + 3(5) = 6 + 15 = 21 \leq 21$$ (satisfied)
- $$5y_1 + 7y_2 = 5(3) + 7(5) = 15 + 35 = 50 \leq 50$$ (satisfied)
6. **Conclusion:**
- The optimal solution of the dual problem is:
$$y_1 = 3$$
$$y_2 = 5$$
- The maximum value of $$P$$ is:
$$P = 363$$
**Final answer:**
$$\boxed{y_1 = 3, \quad y_2 = 5, \quad \max P = 363}$$