Question

Solve the following linear programming problem graphically :
Maximize \( Z = 4500x + 5000y \)
Subject to constraints
\( x + y \leq 250 \)
\( 25x + 40y \leq 7000 \)
\( x \geq 0, y \geq 0 \)

Hint:

Always double-check intersection points by substituting them back into both original equations.
Ensure the non-negativity constraints are satisfied by keeping the region in the first quadrant.

Answer 1

Concept:
• Linear Programming involves identifying a feasible region bounded by linear inequalities.
• Corner Point Theorem states that the optimal value of the objective function occurs at one of the vertices of the feasible region.

Step 1: Convert inequalities to equations and find intercepts
For \( L_1 \): \( x + y = 250 \). Intercepts are \( (250, 0) \) and \( (0, 250) \).
For \( L_2 \): \( 25x + 40y = 7000 \implies 5x + 8y = 1400 \).
At \( x = 0 \), \( 8y = 1400 \implies y = 175 \). Point: \( (0, 175) \).
At \( y = 0 \), \( 5x = 1400 \implies x = 280 \). Point: \( (280, 0) \).

Step 2: Determine the point of intersection
Solving \( x + y = 250 \) and \( 5x + 8y = 1400 \):
From the first equation, \( x = 250 - y \).
Substitute into the second: \( 5(250 - y) + 8y = 1400 \)
\[ 1250 - 5y + 8y = 1400 \implies 3y = 150 \implies y = 50 \]
Then \( x = 250 - 50 = 200 \). Intersection point is \( (200, 50) \).

Step 3: Identify the corner points of the feasible region
Since both constraints are \( \leq \), the region is towards the origin.
The vertices of the feasible region are: \( O(0,0), A(250, 0), B(200, 50), C(0, 175) \).

Step 4: Evaluate \( Z \) at each corner point
\includegraphics[width=1\linewidth]{Q31_sol.png}
• At \( O(0,0) \): \( Z = 4500(0) + 5000(0) = 0 \)
• At \( A(250,0) \): \( Z = 4500(250) + 5000(0) = 1,125,000 \)
• At \( B(200,50) \): \( Z = 4500(200) + 5000(50) = 900,000 + 250,000 = 1,150,000 \)
• At \( C(0,175) \): \( Z = 4500(0) + 5000(175) = 875,000 \)
The maximum value of \( Z \) is \( 1,150,000 \) at \( (200, 50) \).