Max Z = 4x1 − 2x2 Subject to: x1 + x2 <= 14, 5x1 + 2x2 >= 36, 8x1 + 2x2 >= 24, x1 + 10x2 >= 0

To do this, we treat each inequality as an equality and find two points on the line.

• For $x_1 + x_2 = 14$:

  • If $x_1 = 0$, then $x_2 = 14$. Point: $(0, 14)$
  • If $x_2 = 0$, then $x_1 = 14$. Point: $(14, 0)$

• For $5x_1 + 2x_2 = 36$:

  • If $x_1 = 0$, then $2x_2 = 36 \implies x_2 = 18$. Point: $(0, 18)$
  • If $x_2 = 0$, then $5x_1 = 36 \implies x_1 = \frac{36}{5} = 7.2$. Point: $(7.2, 0)$

• For $4x_1 + x_2 = 12$:

  • If $x_1 = 0$, then $x_2 = 12$. Point: $(0, 12)$
  • If $x_2 = 0$, then $4x_1 = 12 \implies x_1 = 3$. Point: $(3, 0)$

We use the inequality signs to determine the correct side of each line.

• $x_1 + x_2 \le 14$: The region below or to the left of the line $x_1 + x_2 = 14$. (Test $(0,0)$: $0+0 \le 14$, True)
• $5x_1 + 2x_2 \ge 36$: The region above or to the right of the line $5x_1 + 2x_2 = 36$. (Test $(0,0)$: $0+0 \ge 36$, False)
• $4x_1 + x_2 \ge 12$: The region above or to the right of the line $4x_1 + x_2 = 12$. (Test $(0,0)$: $0+0 \ge 12$, False)
• $x_1 \ge 0, x_2 \ge 0$: The first quadrant.

Upon plotting, we observe that the constraint $4x_1 + x_2 \ge 12$ is redundant because any point satisfying $5x_1 + 2x_2 \ge 36$ and $x_1 \ge 0, x_2 \ge 0$ will also satisfy $4x_1 + x_2 \ge 12$. Thus, the feasible region is defined by $x_1 + x_2 \le 14$, $5x_1 + 2x_2 \ge 36$, $x_1 \ge 0$, and $x_2 \ge 0$.

The feasible region is a polygon formed by the intersection of the active constraints. The vertices are:

• Vertex A: Intersection of $x_1 + x_2 = 14$ and $5x_1 + 2x_2 = 36$. From $x_1 + x_2 = 14$, we get $x_2 = 14 - x_1$. Substitute into the second equation:

  $$5x_1 + 2(14 - x_1) = 36$$ $$5x_1 + 28 - 2x_1 = 36$$ $$3x_1 = 8 \implies x_1 = \frac{8}{3}$$ $$x_2 = 14 - \frac{8}{3} = \frac{42 - 8}{3} = \frac{34}{3}$$

  So, $A = \left(\frac{8}{3}, \frac{34}{3}\right)$.

• Vertex B: Intersection of $x_1 + x_2 = 14$ and $x_2 = 0$.

  $$x_1 + 0 = 14 \implies x_1 = 14$$

  So, $B = (14, 0)$.

• Vertex C: Intersection of $5x_1 + 2x_2 = 36$ and $x_2 = 0$.

  $$5x_1 + 2(0) = 36 \implies 5x_1 = 36 \implies x_1 = \frac{36}{5} = 7.2$$

  So, $C = \left(\frac{36}{5}, 0\right)$.

The objective function is $Z = 4x_1 - 2x_2$.

• At $A = \left(\frac{8}{3}, \frac{34}{3}\right)$:

  $$Z_A = 4\left(\frac{8}{3}\right) - 2\left(\frac{34}{3}\right) = \frac{32}{3} - \frac{68}{3} = -\frac{36}{3} = -12$$

• At $B = (14, 0)$:

  $$Z_B = 4(14) - 2(0) = 56 - 0 = 56$$

• At $C = \left(\frac{36}{5}, 0\right)$:

  $$Z_C = 4\left(\frac{36}{5}\right) - 2(0) = \frac{144}{5} = 28.8$$

Comparing the values of $Z$ at the corner points, the maximum value is 56.

The maximum value of $Z$ is $\boxed{\text{56}}$ at $x_1 = 14, x_2 = 0$.