Question

The minimum value of the objective function $z = 4x + 6y$ subject to $x + 2y \ge 80$, $3x + y \ge 75$, $x, y \ge 0$ is

• A. 324
• B. 250
• C. 320
• D. 254

Hint:

When evaluating corner points for a minimization problem with non-zero coefficients, the optimal solution is very frequently located at the intersection point of the interior constraint lines rather than the outer axis intercepts. Testing the intersection point $(14,33)$ first can save valuable calculation steps!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question asks for the minimum value of a linear objective function $z$ within a convex feasible region defined by a set of linear inequality constraints.

Step 2: Key Formula or Approach:
According to the Corner Point Theorem of Linear Programming Problems (LPP), the optimal (minimum or maximum) value of an objective function always occurs at one of the corner vertices of the bounded or unbounded feasible region. We need to find the vertices and evaluate $z = 4x + 6y$ at each point.

Step 3: Detailed Explanation:
Let's treat the boundary lines as equalities to find their coordinate axis intercepts:
1. For line $L_1: x + 2y = 80$:
If $x = 0 \implies y = 40 \implies (0, 40)$
If $y = 0 \implies x = 80 \implies (80, 0)$
2. For line $L_2: 3x + y = 75$:
If $x = 0 \implies y = 75 \implies (0, 75)$
If $y = 0 \implies x = 25 \implies (25, 0)$
Since both inequality conditions are of the type "$\ge$", the shaded feasible region is an unbounded zone extending away from the origin. The bounding corner corner-points are:
$A(80, 0)$
$C(0, 75)$
$B$: The intersection point of lines $L_1$ and $L_2$. Let's solve them simultaneously:
Multiply $L_2$ by 2:
$$6x + 2y = 150$$ Subtract $$L_1 \implies (6x + 2y) - (x + 2y) = 150 - 80$$ $$5x = 70 \implies x = 14$$ Substitute $x = 14$ into $L_1$: $14 + 2y = 80 \implies 2y = 66 \implies y = 33$. So, the intersection vertex is $B(14, 33)$.
Now, evaluate the objective cost function $z = 4x + 6y$ at each vertex:
At $A(80, 0)$: $z = 4(80) + 6(0) = 320$
At $B(14, 33)$: $z = 4(14) + 6(33) = 56 + 198 = 254$
At $C(0, 75)$: $z = 4(0) + 6(75) = 450$
Comparing these values, the minimum value obtained is 254 at point $B$. This matches option (D).

Step 4: Final Answer:
The minimum value of the objective function is 254, which corresponds to option (D).