Algebra

Linear Programming

Medium

common

Consider an LPP: Maximise $Z = 50x + 15y$ subjected to constraints

$x + y \leq 60$ ,

$5x + y \leq 100$ ,

$x, y \geq 0$ .

If the maximum value of $Z$ occurs at $x = \alpha$ and $y = \beta$ , then the value of $\alpha + \beta$ is

✅ Correct Option: 2

In any LPP, the maximum or minimum value occurs at a corner point of the feasible region.

The corner points are:

Origin: $(0, 0)$

Where $5x + y = 100$ meets the x-axis, set $y = 0$ :

$5x = 100$

$x = 20$

Point: $(20, 0)$

Where $x + y = 60$ meets the y-axis, set $x = 0$ :

$y = 60$

Point: $(0, 60)$

Intersection of the two lines:

$x + y = 60$ ... (1)

$5x + y = 100$ ... (2)

Subtracting (1) from (2):

$(5x + y) - (x + y) = 100 - 60$

$4x = 40$

$x = 10$

Substituting $x = 10$ in equation (1):

$10 + y = 60$

$y = 50$

Point: $(10, 50)$

Evaluating $Z = 50x + 15y$ at each corner point:

At $(0, 0)$ : $Z = 50(0) + 15(0) = 0$

At $(20, 0)$ : $Z = 50(20) + 15(0) = 1000$

At $(0, 60)$ : $Z = 50(0) + 15(60) = 900$

At $(10, 50)$ : $Z = 50(10) + 15(50) = 500 + 750 = 1250$

The maximum value of $Z$ is $1250$ at point $(10, 50)$ .

Therefore, $\alpha = 10$ and $\beta = 50$

$\alpha + \beta = 10 + 50 = 60$

2025: 27 May Shift 1

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2026: 19 May Shift 2

Two linear programming problems (LPPs) have same bounded feasible region with three corner points $(0, 3), (4, 4), (7, 0)$ . If $Z_1 = x - 3y$ and $Z_2 = 2x + y$ are objective functions of LPPs, then which of the following is/are correct?

A. only $Z_1$ maximizes at $(7, 0)$ .

B. both $Z_1, Z_2$ minimize at $(0, 3)$ .

C. both $Z_1, Z_2$ maximize at $(7, 0)$ .

D. $Z_2$ maximizes at $(4, 4)$ .

Choose the correct answer from the options given below:

2025: 30 May Shift 1

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Solution

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