Problem Statement
Solve the following linear programming problem graphically:
$$\text{Maximize } Z = 4x + 3y$$
subject to the constraints:
$$x + y \le 50$$
$$3x + y \le 90$$
$$x \ge 0, \quad y \ge 0$$
Verified Solution & Marking Scheme
Find boundary lines and intercepts
Line 1: $x + y = 50$. When $x = 0, y = 50$; when $y = 0, x = 50$. Points: $(0, 50), (50, 0)$. Test $(0,0)$: $0 \le 50$ (True, region includes origin).
Line 2: $3x + y = 90$. When $x = 0, y = 90$; when $y = 0, x = 30$. Points: $(0, 90), (30, 0)$. Test $(0,0)$: $0 \le 90$ (True, region includes origin).
Find point of intersection of boundary lines
Subtracting Line 1 from Line 2:
$$(3x + y) - (x + y) = 90 - 50 \implies 2x = 40 \implies x = 20$$
Substitute $x = 20$ in Line 1: $20 + y = 50 \implies y = 30$.
Point of intersection: $B(20, 30)$.
Identify corner points of feasible region
The bounded feasible region $OABC$ has corner points:
$$O(0, 0), \quad A(30, 0), \quad B(20, 30), \quad C(0, 50)$$
Evaluate objective function Z = 4x + 3y at each corner
At $O(0, 0)$: $Z = 4(0) + 3(0) = 0$
At $A(30, 0)$: $Z = 4(30) + 3(0) = 120$
At $B(20, 30)$: $Z = 4(20) + 3(30) = 80 + 90 = 170$
At $C(0, 50)$: $Z = 4(0) + 3(50) = 150$
Hence, the maximum value of $Z$ is $170$ at $x = 20, y = 30$.