Question

The objective function $z = 4x + 5y$ subject to $2x + y \geq 7;\ 2x + 3y \leq 15;\ y \leq 3,\ x \geq 0;\ y \geq 0$ has its minimum value at a point located

• A. on the line $2x + 3y = 15$
• B. on the $X$-axis
• C. on the $Y$-axis
• D. at the origin

Hint:

Since both coefficients in the objective function $z = 4x + 5y$ are positive, to minimize $z$, we want to pull towards the origin as much as possible.
The lower boundary constraint closest to the origin is $2x + y = 7$. Setting $y=0$ minimizes the $y$ contribution entirely and gives $x = 3.5$, yielding $z = 14$. Trying to use the other boundary intersection $D(2,3)$ introduces a large $y$ penalty ($5 \times 3 = 15$), confirming that staying flat on the $X$-axis is the best move.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The problem presents a linear programming problem (LPP). We need to determine where the objective function $z = 4x + 5y$ reaches its minimum value within the feasible region defined by the given system of inequalities.

Step 2: Key Formula or Approach:
According to the Corner Point Theorem, the optimal value (minimum or maximum) of an objective function always occurs at one of the vertices (corner points) of the bounded feasible region. We find the coordinates of these corner points by solving the boundary lines intersecting pairs and evaluate $z$ at each vertex.

Step 3: Detailed Explanation:
Let's find the relevant boundary intersection points:
1. Line $2x + y = 7$ crosses the $X$-axis ($y=0$) at $A\left(\frac{7}{2}, 0\right)$.
2. Line $2x + 3y = 15$ crosses the $X$-axis ($y=0$) at $B\left(\frac{15}{2}, 0\right)$.
3. Intersection of line $2x + 3y = 15$ and line $y = 3$:
$$2x + 3(3) = 15 \implies 2x + 9 = 15 \implies 2x = 6 \implies x = 3 \implies C(3, 3)$$ 4. Intersection of line $2x + y = 7$ and line $y = 3$:
$$2x + 3 = 7 \implies 2x = 4 \implies x = 2 \implies D(2, 3)$$ The shaded feasible region is a quadrilateral bounded by vertices $A\left(\frac{7}{2}, 0\right)$, $B\left(\frac{15}{2}, 0\right)$, $C(3, 3)$, and $D(2, 3)$.
Now, let's evaluate the objective function $z = 4x + 5y$ at each of these four vertices:
$z(A) = 4\left(\frac{7}{2}\right) + 5(0) = 14 + 0 = 14$
$z(B) = 4\left(\frac{15}{2}\right) + 5(0) = 30 + 0 = 30$
$z(C) = 4(3) + 5(3) = 12 + 15 = 27$
$z(D) = 4(2) + 5(3) = 8 + 15 = 23$
Comparing the results ($14, 30, 27, 23$), the minimum value of $z$ is $14$, which occurs at corner point $A\left(\frac{7}{2}, 0\right)$. Since point $A$ has a $y$-coordinate of 0, it lies directly on the $X$-axis.

Step 4: Final Answer:
The objective function reaches its minimum value at a point on the $X$-axis, which matches option (B).