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AMU MCA Linear Programming PYQ

AMU MCA PYQ
The following LPP

Maximize   $z = x_1 + x_2$

Subject to

$x_1 + x_2 \le 1$

$-3x_1 + x_2 \ge 3$

$x_1, x_2 \ge 0$






Go to Discussion
AMU MCA Previous Year PYQ AMU MCA AMU MCA 2020 PYQ

Solution

From $-3x_1 + x_2 \ge 3$

$x_2 \ge 3 + 3x_1$

But from $x_1 + x_2 \le 1$

$x_2 \le 1 - x_1$

So we need:

$3 + 3x_1 \le 1 - x_1$

$4x_1 \le -2$

$x_1 \le -\frac{1}{2}$

This contradicts $x_1 \ge 0$

Hence problem is infeasible.

AMU MCA PYQ
he following LPP:
Maximize $Z = 6x_1 - 2x_2$

Subject to:
$2x_1 - x_2 \le 2$
$x_1 \le 3$
$x_1, x_2 \ge 0$
has optimal value as





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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2022 PYQ

Solution

AMU MCA PYQ
The following LPP:
Minimize $Z = x - y$

Subject to:
$2x + 3y \le 6$
$0 \le x \le 4$
$0 \le y \le 3$

then the number of extreme points on the feasible region and the number of basic feasible solutions are





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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2022 PYQ

Solution

AMU MCA PYQ
A tie for leaving variables in simplex procedure implies





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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2022 PYQ

Solution

AMU MCA PYQ
Optimal value of the following LPP:
Max $z = 2x_1 + 3x_2$
Subject to
       $6x_1 + 5x_2 \le 25$
       $x_1 + 3x_2 \le 10$
        $x_1, x_2 \ge 0$





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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2025 PYQ

Solution

Corner points of feasible region are checked. Maximum value of $z$ occurs at intersection of $6x_1 + 5x_2 = 25$ and $x_1 + 3x_2 = 10$. Substituting gives $x_1 = 2.5$, $x_2 = 3$. $z = 2(2.5) + 3(3) = 5 + 9 = 13.5$.
AMU MCA PYQ
Max $z = 6x_1 - x_2$
Subject to
$2x_1 - x_2 \le 2$
$x_1 \le 3$
$x_1, x_2 \ge 0$
The above LPP has:





Go to Discussion
AMU MCA Previous Year PYQ AMU MCA AMU MCA 2025 PYQ

Solution

The feasible region allows $z$ to increase indefinitely, hence the objective function is unbounded.
AMU MCA PYQ
When 40% of a number is added to 42, the result is the number itself. The number is:





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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2024 PYQ

Solution

Let the number $= x$ $0.4x + 42 = x$ $x - 0.4x = 42$ $0.6x = 42$ $x = 70$
AMU MCA PYQ
The population of a town increases by 5% annually. If it is 15,435 now, its population 2 years ago was:





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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2024 PYQ

Solution

Present population $= 15435$ $P(1.05)^2 = 15435$ $P \times 1.1025 = 15435$ $P = \frac{15435}{1.1025} = 14000$

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