Problem Statement
Solve the following Linear Programming Problem graphically:
$\text{Maximize } Z = 250x + 70y$
subject to the constraints:
$\begin{aligned} 3x + y &\le 66 \\ x + y &\le 45 \\ x &\ge 0, \quad y \ge 0 \end{aligned}$
Verified Solution & Marking Scheme
Convert Inequalities to Boundary Lines and Find Intercepts
1. Line $L_1$: $3x + y = 66$
- At $x = 0$, $y = 66 \implies (0, 66)$
- At $y = 0$, $3x = 66 \implies x = 22 \implies (22, 0)$
- Testing $(0, 0)$: $3(0) + 0 = 0 \le 66$ (True, region contains origin)
2. Line $L_2$: $x + y = 45$
- At $x = 0$, $y = 45 \implies (0, 45)$
- At $y = 0$, $x = 45 \implies (45, 0)$
- Testing $(0, 0)$: $0 + 0 = 0 \le 45$ (True, region contains origin)
Determine Intersection Point of Boundary Lines
Subtracting $x + y = 45$ from $3x + y = 66$:
$(3x + y) - (x + y) = 66 - 45$
$2x = 21 \implies x = 10.5$
Substituting $x = 10.5$ into $x + y = 45$:
$y = 45 - 10.5 = 34.5$
Intersection point $B$ is $(10.5, 34.5)$.
Identify Corner Points of the Feasible Region
The feasible region is bounded by the polygon $OABC$ with corner points:
- $O(0, 0)$
- $A(22, 0)$ (x-intercept of $3x + y = 66$)
- $B(10.5, 34.5)$ (intersection of $L_1$ and $L_2$)
- $C(0, 45)$ (y-intercept of $x + y = 45$)
Evaluate Objective Function Z at Each Corner Point
- At $O(0, 0)$: $Z = 250(0) + 70(0) = 0$
- At $A(22, 0)$: $Z = 250(22) + 70(0) = 5500$
- At $B(10.5, 34.5)$: $Z = 250(10.5) + 70(34.5) = 2625 + 2415 = 5040$
- At $C(0, 45)$: $Z = 250(0) + 70(45) = 3150$
The maximum value of $Z$ is $5500$ at the point $(22, 0)$.