Question

The objective function z = x1 + x2 subject to the conditions are

x1 + x2 ≤ 10, -2x1 + 3x2 ≤ 15, x1 ≤ 6, x1, x2 ≥ 0

has maximum value -------------of the feasible region

• A. At only one point
• B. at only two point
• C. at every point of the segment joining two point
• D. at every point of the line joining two points equivalent to

Answer 1

Answer: (C)

at every point of the segment joining two point

Answer 2

To find the maximum value of the objective function $z = x_1 + x_2$ subject to the given constraints, we first need to determine the feasible region defined by these inequalities.

The constraints are:

1. $x_1 + x_2 \le 10$
2. $-2x_1 + 3x_2 \le 15$
3. $x_1 \le 6$
4. $x_1 \ge 0$
5. $x_2 \ge 0$

Step 1: Graph the inequalities and find the feasible region.

• $x_1 + x_2 \le 10$: Draw the line $x_1 + x_2 = 10$. It passes through (10, 0) and (0, 10). The region satisfying the inequality is below this line (towards the origin).
• $-2x_1 + 3x_2 \le 15$: Draw the line $-2x_1 + 3x_2 = 15$. It passes through (0, 5) and (-7.5, 0). The region satisfying the inequality is above this line (towards the origin).
• $x_1 \le 6$: Draw the vertical line $x_1 = 6$. The region satisfying the inequality is to the left of this line.
• $x_1 \ge 0, x_2 \ge 0$: This restricts the feasible region to the first quadrant.

The feasible region is a polygon formed by the intersection of all these regions.

Step 2: Identify the corner points (vertices) of the feasible region.

The corner points are the intersections of the boundary lines that satisfy all constraints.

1. O (0, 0): Intersection of $x_1=0$ and $x_2=0$. All constraints are satisfied. $z = 0+0 = 0$.

2. A (0, 5): Intersection of $x_1=0$ and $-2x_1 + 3x_2 = 15$. Substitute $x_1=0$ into $-2x_1 + 3x_2 = 15 \Rightarrow 3x_2 = 15 \Rightarrow x_2 = 5$. Check constraints: $0+5 \le 10$ (True), $0 \le 6$ (True). $z = 0+5 = 5$.

3. B (6, 0): Intersection of $x_1=6$ and $x_2=0$. Check constraints: $6+0 \le 10$ (True), $-2(6)+3(0) = -12 \le 15$ (True). $z = 6+0 = 6$.

4. C (6, 4): Intersection of $x_1=6$ and $x_1 + x_2 = 10$. Substitute $x_1=6$ into $x_1 + x_2 = 10 \Rightarrow 6 + x_2 = 10 \Rightarrow x_2 = 4$. Check constraints: $-2(6)+3(4) = -12+12 = 0 \le 15$ (True). $z = 6+4 = 10$.

5. E (3, 7): Intersection of $x_1 + x_2 = 10$ and $-2x_1 + 3x_2 = 15$. From $x_1 + x_2 = 10$, we get $x_1 = 10 - x_2$. Substitute into $-2x_1 + 3x_2 = 15$: $-2(10 - x_2) + 3x_2 = 15$ $-20 + 2x_2 + 3x_2 = 15$ $5x_2 = 35 \Rightarrow x_2 = 7$. Then $x_1 = 10 - 7 = 3$. Check constraints: $3 \le 6$ (True). $z = 3+7 = 10$.

The vertices of the feasible region are O(0,0), A(0,5), E(3,7), C(6,4), and B(6,0).

Step 3: Evaluate the objective function $z = x_1 + x_2$ at each vertex.

• At O(0, 0): $z = 0 + 0 = 0$
• At A(0, 5): $z = 0 + 5 = 5$
• At B(6, 0): $z = 6 + 0 = 6$
• At C(6, 4): $z = 6 + 4 = 10$
• At E(3, 7): $z = 3 + 7 = 10$

Step 4: Determine the maximum value and where it occurs.

The maximum value of $z$ is 10. This maximum value occurs at two distinct vertices: C(6, 4) and E(3, 7).

According to the fundamental theorem of linear programming, if the optimal value of the objective function occurs at two distinct corner points of the feasible region, then it occurs at every point on the line segment joining these two points. In this case, the objective function $z = x_1 + x_2$ is parallel to the constraint line $x_1 + x_2 = 10$, which forms the edge CE of the feasible region. Therefore, every point $(x_1, x_2)$ on the line segment joining C(6, 4) and E(3, 7) will yield $x_1 + x_2 = 10$, resulting in the maximum value of $z=10$.

The maximum value occurs at every point of the segment joining the two points C(6,4) and E(3,7).