JAMIA 2023 — Mathematics PYQ

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Question

JAMIA 2023 — Mathematics PYQ

JAMIA | Mathematics | 2023

Z=7x+y, subject to constraints:
5𝑥+𝑦≥5,
𝑥+𝑦≥3
𝑥≥0,𝑦≥0,𝑦
Then minimum value of Z occurs at:

Choose the correct answer:

Explanation

1. Objective Function and Constraints

Minimize: $Z = 7x + y$
Constraints:
1. $5x + y \geq 5$
2. $x + y \geq 3$
3. $x \geq 0, y \geq 0$

2. Identify Boundary Lines and Intersections

We find the intersection points of the boundary lines with the axes and with each other.

Line $L_1: 5x + y = 5$

When $x = 0$ , $y = 5 \implies (0, 5)$
When $y = 0$ , $x = 1 \implies (1, 0)$

Line $L_2: x + y = 3$

When $x = 0$ , $y = 3 \implies (0, 3)$
When $y = 0$ , $x = 3 \implies (3, 0)$

Intersection of $L_1$ and $L_2$ :

Solve the system:

$5x + y = 5$
$x + y = 3 \implies y = 3 - x$

Substitute $y$ in the first equation:

5x + (3 - x) = 5

4x = 2 \implies x = \frac{1}{2}

y = 3 - \frac{1}{2} = \frac{5}{2}

Intersection point: $(\frac{1}{2}, \frac{5}{2})$ or $(0.5, 2.5)$ .

3. Determine the Feasible Region Corner Points

The constraints are "greater than or equal to," so the feasible region is unbounded and lies above the lines. We test the points to see which ones bound the feasible region:

$(0, 5)$ : Satisfies both $5x+y \geq 5$ and $x+y \geq 3$ .
$(\frac{1}{2}, \frac{5}{2})$ : Satisfies both (intersection point).
$(3, 0)$ : Satisfies both $5x+y \geq 5$ and $x+y \geq 3$ .

The corner points are: $(0, 5)$ , $(0.5, 2.5)$ , and $(3, 0)$ .

4. Calculate $Z$ at Corner Points

We substitute the corner points into $Z = 7x + y$ :

At $(0, 5)$ :

$Z = 7(0) + 5 = 5$
At $(0.5, 2.5)$ :

$Z = 7(0.5) + 2.5 = 3.5 + 2.5 = 6$
At $(3, 0)$ :

$Z = 7(3) + 0 = 21$

5. Final Answer

Comparing the values $5, 6,$ and $21$ , the minimum value is $5$ .

Result:

The minimum value of $Z$ occurs at:

(0, 5)