Exercise 12.1 Practice
Chapter 12: Linear Programming – NCERT Solutions
Q1
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Maximize $Z = 3x + 4y$ subject to the constraints: $x + y \le 4, x \ge 0, y \ge 0$.
The feasible region is bounded by the triangle with corner points $(0,0), (4,0), (0,4)$.
Evaluate $Z$ at corner points:
At $(0,0): Z = 0$.
At $(4,0): Z = 3(4) + 0 = 12$.
At $(0,4): Z = 0 + 4(4) = 16$.
$\boxed{\text{Maximum } Z = 16 \text{ at } (0,4)}$
Q2
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Minimize $Z = -3x + 4y$ subject to $x + 2y \le 8, 3x + 2y \le 12, x \ge 0, y \ge 0$.
Corner points of the feasible region are $(0,0), (4,0), (2,3), (0,4)$.
$Z(0,0) = 0$.
$Z(4,0) = -3(4) + 0 = -12$.
$Z(2,3) = -3(2) + 4(3) = 6$.
$Z(0,4) = 16$.
$\boxed{\text{Minimum } Z = -12 \text{ at } (4,0)}$
Q3
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Maximize $Z = 5x + 3y$ subject to $3x + 5y \le 15, 5x + 2y \le 10, x \ge 0, y \ge 0$.
Corner points: $(0,0), (2,0), (\frac{20}{19}, \frac{45}{19}), (0,3)$.
$Z(0,0) = 0$.
$Z(2,0) = 10$.
$Z(0,3) = 9$.
$Z(\frac{20}{19}, \frac{45}{19}) = 5(\frac{20}{19}) + 3(\frac{45}{19}) = \frac{235}{19} \approx 12.37$.
$\boxed{\text{Maximum } Z = \frac{235}{19} \text{ at } (\frac{20}{19}, \frac{45}{19})}$
Q4
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Minimize $Z = 3x + 5y$ such that $x + 3y \ge 3, x + y \ge 2, x, y \ge 0$.
Unbounded region corner points: $(3,0), (1.5, 0.5), (0,2)$.
$Z(3,0) = 9$.
$Z(1.5, 0.5) = 3(1.5) + 5(0.5) = 4.5 + 2.5 = 7$.
$Z(0,2) = 10$.
$\boxed{\text{Minimum } Z = 7 \text{ at } (1.5, 0.5)}$
Q5
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Maximize $Z = 3x + 2y$ subject to $x + 2y \le 10, 3x + y \le 15, x, y \ge 0$.
Corner points: $(0,0), (5,0), (4,3), (0,5)$.
$Z(5,0) = 15$.
$Z(4,3) = 3(4) + 2(3) = 18$.
$Z(0,5) = 10$.
$\boxed{\text{Maximum } Z = 18 \text{ at } (4,3)}$
Q6
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Minimize $Z = x + 2y$ subject to $2x + y \ge 3, x + 2y \ge 6, x, y \ge 0$.
Corner points of unbounded region: $(6,0)$ and $(0,3)$.
$Z(6,0) = 6$.
$Z(0,3) = 6$.
$\boxed{\text{Minimum } Z = 6 \text{ at all points on the segment joining } (6,0) \text{ and } (0,3)}$
Q7
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Minimize and Maximize $Z = 5x + 10y$ subject to $x + 2y \le 120, x + y \ge 60, x - 2y \ge 0, x, y \ge 0$.
Corner points: $(60,0), (120,0), (60,30), (40,20)$.
$Z(60,0) = 300$ (Min).
$Z(120,0) = 600$ (Max).
$Z(60,30) = 300 + 300 = 600$ (Max).
$Z(40,20) = 200 + 200 = 400$.
$\boxed{\text{Min } Z = 300 \text{ at } (60,0); \text{ Max } Z = 600 \text{ at } (120,0) \text{ and } (60,30)}$
Q8
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Minimize and Maximize $Z = x + 2y$ subject to $x + 2y \ge 100, 2x - y \le 0, 2x + y \le 200, x, y \ge 0$.
Corner points: $(0,50), (0,200), (50,100), (20,40)$.
$Z(0,50) = 100$ (Min).
$Z(0,200) = 400$ (Max).
$Z(50,100) = 250$.
$Z(20,40) = 100$ (Min).
$\boxed{\text{Min } Z = 100 \text{ at } (0,50) \text{ and } (20,40); \text{ Max } Z = 400 \text{ at } (0,200)}$
Q9
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Maximize $Z = -x + 2y$ subject to $x \ge 3, x + y \ge 5, x + 2y \ge 6, y \ge 0$.
The feasible region is unbounded. Corner points: $(6,0), (4,1), (3,2)$.
$Z(6,0) = -6, Z(4,1) = -2, Z(3,2) = 1$.
Since the region is unbounded, we check if $-x + 2y > 1$ is possible.
The half plane $-x + 2y > 1$ has common points with the feasible region. Thus, $Z$ can be arbitrarily large.
$\boxed{\text{No maximum value}}$
Q10
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Maximize $Z = x + y$ subject to $x - y \le -1, -x + y \le 0, x, y \ge 0$.
Constraints: $y \ge x + 1$ and $y \le x$.
These two conditions ($y \ge x+1$ and $y \le x$) are contradictory and cannot be satisfied simultaneously.
There is no feasible region.
$\boxed{\text{No maximum value}}$