Question: Solve the given linear programming problem using the table method. A table of basic solutions is provided to the right.
Maximize $P = 24x_1 + 10x_2$
$5x_1 + 16x_2 \leq 140$
$x_1 + 3x_2 \leq 27$
$x_1, x_2 \geq 0$
$x_1 \quad x_2 \quad s_1 \quad s_2$
$0 \quad 0 \quad 140 \quad 27 \quad$ Feasible
$0 \quad 8.75 \quad 0 \quad 0.75 \quad$ Feasible
$0 \quad 9 \quad -4 \quad 0 \quad$ Not feasible
$28 \quad 0 \quad 0 \quad -1 \quad$ Not feasible
$27 \quad 0 \quad 5 \quad 0 \quad$ Feasible
$12 \quad 5 \quad 0 \quad 0 \quad$ Feasible
The maximum value of $P$ is $P = __$, occuring at $x_1 = __$, $x_2 = __$.
1. **State the problem:**
We want to maximize the objective function $$P = 24x_1 + 10x_2$$
subject to the constraints:
$$5x_1 + 16x_2 \leq 140$$
$$x_1 + 3x_2 \leq 27$$
$$x_1, x_2 \geq 0$$
2. **Understand the table of basic solutions:**
The table lists candidate solutions $(x_1, x_2)$ with slack variables $s_1, s_2$ and feasibility status.
3. **Identify feasible solutions:**
From the table, feasible solutions are:
- $(0, 0)$ with $P = 24(0) + 10(0) = 0$
- $(0, 8.75)$ with $P = 24(0) + 10(8.75) = 87.5$
- $(27, 0)$ with $P = 24(27) + 10(0) = 648$
- $(12, 5)$ with $P = 24(12) + 10(5) = 288 + 50 = 338$
4. **Evaluate $P$ at each feasible point:**
- At $(0,0)$: $$P = 0$$
- At $(0,8.75)$: $$P = 10 \times 8.75 = 87.5$$
- At $(27,0)$: $$P = 24 \times 27 = 648$$
- At $(12,5)$: $$P = 24 \times 12 + 10 \times 5 = 288 + 50 = 338$$
5. **Find the maximum value:**
The maximum $P$ is $$648$$ at $$x_1 = 27, x_2 = 0$$.
**Final answer:**
The maximum value of $P$ is $\boxed{648}$, occurring at $x_1 = \boxed{27}$, $x_2 = \boxed{0}$.