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: Understand the problem.
We minimise $z = 4x + 5y$ subject to $2x + y \ge 7$, $2x + 3y \le 15$, $y \le 3$, $x \ge 0$, $y \ge 0$.
Step 2: Recall the corner-point rule.
The smallest or largest value of $z$ in a linear program happens at a corner of the feasible region. So we find the corners and test $z$ at each.
Step 3: Find corner points on the $x$-axis.
$2x + y = 7$ meets $y = 0$ at $A\left(\tfrac{7}{2}, 0\right)$. $2x + 3y = 15$ meets $y = 0$ at $B\left(\tfrac{15}{2}, 0\right)$.
Step 4: Find corners with $y = 3$.
$2x + 3y = 15$ with $y = 3$: $2x + 9 = 15$, so $x = 3$, giving $C(3,3)$. $2x + y = 7$ with $y = 3$: $2x = 4$, so $x = 2$, giving $D(2,3)$.
Step 5: Evaluate $z$ at each corner.
$z(A) = 4(\tfrac72) + 0 = 14$; $z(B) = 4(\tfrac{15}{2}) = 30$; $z(C) = 12 + 15 = 27$; $z(D) = 8 + 15 = 23$.
Step 6: Pick the minimum.
The smallest is 14 at $A\left(\tfrac72, 0\right)$. Since its $y$-coordinate is 0, this point lies on the $X$-axis. \[ \boxed{\text{on the } X\text{-axis}} \]