Maths MCQ

A set A is convex if, for any two points $X_1, X_2$$\in\text{A}$ and $\lambda\in0,1$ imply that $\lambda\times1+1-\lambda\times2\in\text{A}$.
Since, here $X_1$ and $X_2$​​​​​​​ are optimal solution
Therefore, their convex combination will also be an optimal solution
Thus, $\text{X}=\lambda\ \text{X}_1+(1-\lambda)\text{X}_2,0\leq\lambda\leq1$ gives an optimal solution.

We need to maximize the function $Z = 3x_4 + 5x_2$_
First, we will convert the given inequations into equations, we obtain the following equations:
$3x_1 + 2x_2 = 18, x_1 = 4, x_2 = 6, x_1 = 0$ and $x_2 = 0$
Region represented by $3x_1 + 2x_2 \leq 18:$
The line $3x_1 + 2x_2 = 18$ meets the coordinate axes at $A(6, 0)$ and $B(0, 9)$ respectively.
By joining these points we obtain the line $3X1 + 2x2 = 18.$
Clearly $(0, 0)$ satisfies the inequation $3x_1 + 2x_2 = 18.$
So the region in the plane which contain the origin represents the solution set of the inequation $3x_1 + 2x_2 \leq 18.$
Region represented by $x_1 \leq 4:$
The line $x_1 = 4$ is the line that passes through $C(4, 0)$ and is parallel to the $Y$ axis.
The region to the left of the line $x_1 = 4$ will satisfy the inequation $x_1 \leq 4.$
Region represented by $x_2 \leq 6:$
The line $x_2 = 6$ is the line that passes through $D(0, 6)$ and is parallel to the $X$ axis.
The region below the line $x_2 = 6$ will satisfy the inequation $X_2 \leq 6.$
Region represented by $x_1 \geq 0$ and $x_2 \geq 0:$
Since, every point in the first quadrant satisfies these inequations.
So, the first quadrant is the region represented by the inequations $x_1 \geq 0$ and $x_2 \geq 0.$
The feasible region determined by the system of constraints, $3x_1 + 2x_2 \leq 18, x_1 \leq 4, x_2 \leq 6, x_1 \geq 0$ and $x_2 \geq 0$ are as follows

Corner points are $O(0, 0), D(0, 6), F(2, 6), E(4, 3)$ and $C(4, 0).$
The values of the objective function at these points are given in the following table.