Word problem of Linear programming Questions in English

(C) Step $1$: Identify the corner points of the feasible region defined by the constraints.
Step $2$: The lines are $2x + 3y = 6$ and $x + y = 1$.
Step $3$: Intersection of $2x + 3y = 6$ and $x + y = 1$ gives $x = 3, y = -2$ (outside the first quadrant).
Step $4$: The vertices of the feasible region are $(0, 1)$, $(0, 2)$, $(1, 0)$, and $(3, 0)$.
Step $5$: Evaluate $z = 3x + y$ at each vertex:
At $(0, 1)$, $z = 3(0) + 1 = 1$.
At $(0, 2)$, $z = 3(0) + 2 = 2$.
At $(1, 0)$, $z = 3(1) + 0 = 3$.
At $(3, 0)$, $z = 3(3) + 0 = 9$.
Step $6$: Maximum value is $9$ and minimum value is $1$.
Step $7$: The difference is $9 - 1 = 8$.

(D) For the maximum value of $z$ to occur at both points $(15, 15)$ and $(0, 20)$, the value of $z$ must be equal at these two points.
Substitute the coordinates into the objective function $z = px + qy$:
At $(15, 15)$, $z_1 = p(15) + q(15) = 15p + 15q$.
At $(0, 20)$, $z_2 = p(0) + q(20) = 20q$.
Equating $z_1$ and $z_2$:
$15p + 15q = 20q$.
Subtract $15q$ from both sides:
$15p = 5q$.
Divide by $5$:
$q = 3p$.