Question

Solve the following linear programming problem graphically: Maximize \( Z = 8000x + 12000y \) subject to the boundary constraints: \( 3x + 4y \le 60 \), \( x + 3y \le 30 \), with non-negativity restrictions \( x \ge 0, y \ge 0 \).

Hint:

When graphing lines quickly, always choose coordinates that generate integers for intercepts. If any point is fractional, scale your graph axis units carefully (e.g., choose steps of 5 or 10 units per grid square block) to minimize drawing alignment errors!

Answer 1

Concept: To solve a Linear Programming Problem (LPP) graphically, we complete the following operational steps:
• Treat inequalities as linear equations to plot boundary lines on a Cartesian coordinate grid.
• Determine the valid half-plane region for each inequality (by testing a point like the origin \((0,0)\)).
• Identify the enclosed intersection space known as the Feasible Region.
• Find all specific outer vertices (corner points) of this region.
• Evaluate the objective function \(Z\) at each corner point. The Corner Point Theorem guarantees that the optimal value will occur at one of these vertices.

Step 1: Graph line 1 corresponding to Constraint 1 (\(3x + 4y = 60\)).
Let us find the axial intercept coordinates for this line:
• Set \( x = 0 \): \( 3(0) + 4y = 60 \Rightarrow 4y = 60 \Rightarrow y = 15 \). Point is \( A(0, 15) \).
• Set \( y = 0 \): \( 3x + 4(0) = 60 \Rightarrow 3x = 60 \Rightarrow x = 20 \). Point is \( B(20, 0) \). Test origin \((0,0)\) in \( 3x + 4y \le 60 \): \( 0 \le 60 \), which is true. Hence, the region lies towards the origin side of line 1.

Step 2: Graph line 2 corresponding to Constraint 2 (\(x + 3y = 30\)).
Let us find the intercept coordinates for this second line:
• Set \( x = 0 \): \( 0 + 3y = 30 \Rightarrow 3y = 30 \Rightarrow y = 10 \). Point is \( C(0, 10) \).
• Set \( y = 0 \): \( x + 3(0) = 30 \Rightarrow x = 30 \). Point is \( D(30, 0) \). Test origin \((0,0)\) in \( x + 3y \le 30 \): \( 0 \le 30 \), which is true. Hence, this region also lies towards the origin side of line 2.

Step 3: Solve simultaneously to find the intersection point \(E\) of both lines.
From line 2, express \(x\) as: \[ x = 30 - 3y \quad \cdots (1) \] Substitute this expression into line 1's equation: \[ 3(30 - 3y) + 4y = 60 \] Expand and group variable components: \[ 90 - 9y + 4y = 60 \] \[ 90 - 5y = 60 \quad \Rightarrow \quad 5y = 30 \quad \Rightarrow \quad y = 6 \] Substitute \(y = 6\) back into equation (1) to find \(x\): \[ x = 30 - 3(6) = 30 - 18 = 12 \] Hence, the intersection point of the boundary limits is \( E(12, 6) \).

Step 4: Identify the bounded feasible region and evaluate \(Z\).
Since \(x \ge 0\) and \(y \ge 0\), our search space is restricted exclusively to the first quadrant. Taking the intersection of all half-planes, the vertices of the bounded feasible region are found to be: \[ O(0, 0), \quad C(0, 10), \quad E(12, 6), \quad B(20, 0) \] Let us compute the value of the objective function \( Z = 8000x + 12000y \) at each vertex in a comprehensive table: {|c|c|l|} Corner Point & Coordinates \((x, y)\) & Value of Objective Function \(Z = 8000x + 12000y\)
\(O\) & \((0, 0)\) & \(Z = 8000(0) + 12000(0) = 0\)
\(C\) & \((0, 10)\) & \(Z = 8000(0) + 12000(10) = 120,000\)
\(\mathbf{E}\) & \(\mathbf{(12, 6)}\) & \(\mathbf{Z = 8000(12) + 12000(6) = 96,000 + 72,000 = 168,000}\)
\(B\) & \((20, 0)\) & \(Z = 8000(20) + 12000(0) = 160,000\)
Comparing all calculated values, the maximum value obtained is exactly \(168,000\), which occurs at the coordinate vertex point \((12, 6)\).