CUET UG Mathematics Applied Linear Programming Previous Year Questions (PYQs)
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The maximum value of $z=4x+2y$ subject to constraints
$2x+3y \le 28$,
$x+y \le 10$,
$x,y \ge 0$ is:
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Solution
Corner points of feasible region:
$(0,0)$
From $x+y=10$
If $x=0$, $y=10$ → but check first constraint:
$2(0)+3(10)=30>28$ ❌ not feasible
So take intersection of
$2x+3y=28$
$x+y=10$
Solve:
$y=10-x$
Substitute:
$2x+3(10-x)=28$
$2x+30-3x=28$
$-x=-2$
$x=2$
Then
$y=10-2=8$
So intersection point $(2,8)$
Now check intercepts:
If $y=0$
From $x+y=10$ → $x=10$
Check first constraint:
$2(10)+0=20\le 28$ ✔
So point $(10,0)$ is feasible.
Now evaluate $z=4x+2y$
At $(0,0)$ → $0$
At $(2,8)$ → $4(2)+2(8)=8+16=24$
At $(10,0)$ → $40$
Maximum value = 40
Corner points of the feasible region for an LPP are
$(0,2),; (3,0),; (6,0)$ and $(6,8)$.
If $z=2x+3y$ is the objective function of LPP then
$\max(z)-\min(z)$ is equal to:
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Solution
Evaluate $z=2x+3y$ at each corner point:
At $(0,2)$
$z=2(0)+3(2)=6$
At $(3,0)$
$z=2(3)+3(0)=6$
At $(6,0)$
$z=2(6)+0=12$
At $(6,8)$
$z=2(6)+3(8)=12+24=36$
So,
$\max(z)=36$
$\min(z)=6$
Therefore,
$\max(z)-\min(z)=36-6=30$
The corner points of the feasible region for an L.P.P are
$(2,0),(7,0),(4,5),(0,3)$
and
$z=2x+3y$
is the objective function.
The difference of the maximum and minimum values of $z$ is:
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Solution
Evaluate $z=2x+3y$ at each corner point.
At $(2,0)$
$z=2(2)+3(0)=4$
At $(7,0)$
$z=2(7)+3(0)=14$
At $(4,5)$
$z=2(4)+3(5)=8+15=23$
At $(0,3)$
$z=2(0)+3(3)=9$
Maximum value of $z=23$
Minimum value of $z=4$
Difference
$23-4=19$
The corner points of the feasible region determined by x + y ≤ 8, 2x + y ≥ 8, x ≥ 0, y ≥ 0 are A(0, 8), B(4, 0), and C(8, 0). If the objective function Z = ax + by has its maximum value on the line segment AB, then the relation between a and b is:
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Solution
An objective function Z = ax + by is maximum at points (8, 2) and (4, 6). If
a ≥ 0, b ≥ 0, and ab = 25, then the maximum value of the function is:
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Solution
CUET UG Mathematics Applied
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CUET UG Mathematics Applied
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