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NCERT Solutions · Class 12 · Chapter 12

NCERT Solutions for Class 12 Maths Chapter 12: Linear Programming

Formulating and solving linear programming problems in two variables graphically, by the corner point method, including unbounded and empty feasible regions. Our own step-by-step solutions to every question in Exercise 12.1, set out for step marks.

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Exercise 12.1: Graphical method (corner points)

What it tests. Shade the feasible region (all constraints together, with \(x, y \ge 0\)), find its corner points by solving pairs of boundary lines, and evaluate Z at each. A bounded region always has both a maximum and a minimum at corners. If the region is unbounded, the best corner value M is the answer only if the half-plane \(Z > M\) (for a maximum) or \(Z < M\) (for a minimum) has no point in common with the region. If two adjacent corners tie, every point of the edge between them is optimal.

Exercise 12.1, Question 1

Maximise \(Z = 3x + 4y\) subject to \(x + y \le 4\), \(x \ge 0\), \(y \ge 0\).
Show solution
  1. Draw the constraint lines, shade the feasible region and list its corner points; evaluate Z at each.
  2. \[\begin{array}{c|c}\text{corner} & Z \\ \hline (0, 0) & 0 \\ (4, 0) & 12 \\ (0, 4) & 16\end{array}\]
Answer: Maximum \(Z = 16\) at \((0, 4)\)

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Exercise 12.1, Question 2

Minimise \(Z = -3x + 4y\) subject to \(x + 2y \le 8\), \(3x + 2y \le 12\), \(x \ge 0\), \(y \ge 0\).
Show solution
  1. Draw the constraint lines, shade the feasible region and list its corner points; evaluate Z at each.
  2. \(x + 2y = 8\) and \(3x + 2y = 12\) meet at \((2, 3)\).
  3. \[\begin{array}{c|c}\text{corner} & Z \\ \hline (0, 0) & 0 \\ (4, 0) & -12 \\ (2, 3) & 6 \\ (0, 4) & 16\end{array}\]
Answer: Minimum \(Z = -12\) at \((4, 0)\)

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Exercise 12.1, Question 3

Maximise \(Z = 5x + 3y\) subject to \(3x + 5y \le 15\), \(5x + 2y \le 10\), \(x \ge 0\), \(y \ge 0\).
Show solution
  1. Draw the constraint lines, shade the feasible region and list its corner points; evaluate Z at each.
  2. \(3x + 5y = 15\) and \(5x + 2y = 10\) meet at \[\left(\tfrac{20}{19}, \tfrac{45}{19}\right)\]
  3. \[\begin{array}{c|c}\text{corner} & Z \\ \hline (0, 0) & 0 \\ (2, 0) & 10 \\ \left(\tfrac{20}{19}, \tfrac{45}{19}\right) & \tfrac{235}{19} \\ (0, 3) & 9\end{array}\]
Answer: Maximum \(Z = \tfrac{235}{19}\) at \[\left(\tfrac{20}{19}, \tfrac{45}{19}\right)\]

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Exercise 12.1, Question 4

Minimise \(Z = 3x + 5y\) such that \(x + 3y \ge 3\), \(x + y \ge 2\), \(x, y \ge 0\).
Show solution
  1. Draw the constraint lines, shade the feasible region and list its corner points; evaluate Z at each. The region is unbounded.
  2. \(x + 3y = 3\) and \(x + y = 2\) meet at \(\left(\tfrac32, \tfrac12\right)\).
  3. \[\begin{array}{c|c}\text{corner} & Z \\ \hline (3, 0) & 9 \\ \left(\tfrac32, \tfrac12\right) & 7 \\ (0, 2) & 10\end{array}\]
  4. The half-plane \(3x + 5y < 7\) has no point in common with the region, so 7 is the minimum.
Answer: Minimum \(Z = 7\) at \(\left(\tfrac32, \tfrac12\right)\)

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Exercise 12.1, Question 5

Maximise \(Z = 3x + 2y\) subject to \(x + 2y \le 10\), \(3x + y \le 15\), \(x, y \ge 0\).
Show solution
  1. Draw the constraint lines, shade the feasible region and list its corner points; evaluate Z at each.
  2. \(x + 2y = 10\) and \(3x + y = 15\) meet at \((4, 3)\).
  3. \[\begin{array}{c|c}\text{corner} & Z \\ \hline (0, 0) & 0 \\ (5, 0) & 15 \\ (4, 3) & 18 \\ (0, 5) & 10\end{array}\]
Answer: Maximum \(Z = 18\) at \((4, 3)\)

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Exercise 12.1, Question 6

Minimise \(Z = x + 2y\) subject to \(2x + y \ge 3\), \(x + 2y \ge 6\), \(x, y \ge 0\). Show that the minimum occurs at more than two points.
Show solution
  1. Draw the constraint lines, shade the feasible region and list its corner points; evaluate Z at each. The region is unbounded, with corners \((6, 0)\) and \((0, 3)\) (the line \(2x + y = 3\) meets \(x + 2y = 6\) at \((0, 3)\)).
  2. \[\begin{array}{c|c}\text{corner} & Z \\ \hline (6, 0) & 6 \\ (0, 3) & 6\end{array}\]
  3. \(x + 2y < 6\) has no point in common with the region, so 6 is the minimum; it is reached at every point of the edge \(x + 2y = 6\) between \((6, 0)\) and \((0, 3)\).
Answer: Minimum \(Z = 6\), at every point of the segment joining \((6, 0)\) and \((0, 3)\)

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Exercise 12.1, Question 7

Minimise and maximise \(Z = 5x + 10y\) subject to \(x + 2y \le 120\), \(x + y \ge 60\), \(x - 2y \ge 0\), \(x, y \ge 0\).
Show solution
  1. Draw the constraint lines, shade the feasible region and list its corner points; evaluate Z at each.
  2. Corners: \((60, 0)\), \((120, 0)\), \((60, 30)\) (from \(x + 2y = 120\), \(x = 2y\)), \((40, 20)\) (from \(x + y = 60\), \(x = 2y\)).
  3. \[\begin{array}{c|c}\text{corner} & Z \\ \hline (60, 0) & 300 \\ (120, 0) & 600 \\ (60, 30) & 600 \\ (40, 20) & 400\end{array}\]
Answer: Minimum \(Z = 300\) at \((60, 0)\); maximum \(Z = 600\) at every point of the segment joining \((120, 0)\) and \((60, 30)\)

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Exercise 12.1, Question 8

Minimise and maximise \(Z = x + 2y\) subject to \(x + 2y \ge 100\), \(2x - y \le 0\), \(2x + y \le 200\), \(x, y \ge 0\).
Show solution
  1. Draw the constraint lines, shade the feasible region and list its corner points; evaluate Z at each.
  2. Corners: \((0, 50)\), \((20, 40)\) (from \(x + 2y = 100\), \(y = 2x\)), \((50, 100)\) (from \(y = 2x\), \(2x + y = 200\)), \((0, 200)\).
  3. \[\begin{array}{c|c}\text{corner} & Z \\ \hline (0, 50) & 100 \\ (20, 40) & 100 \\ (50, 100) & 250 \\ (0, 200) & 400\end{array}\]
Answer: Minimum \(Z = 100\) at every point of the segment joining \((0, 50)\) and \((20, 40)\); maximum \(Z = 400\) at \((0, 200)\)

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Exercise 12.1, Question 9

Maximise \(Z = -x + 2y\) subject to \(x \ge 3\), \(x + y \ge 5\), \(x + 2y \ge 6\), \(y \ge 0\).
Show solution
  1. Draw the constraint lines, shade the feasible region and list its corner points; evaluate Z at each. The region is unbounded.
  2. Corners: \((6, 0)\), \((4, 1)\), \((3, 2)\).
  3. \[\begin{array}{c|c}\text{corner} & Z \\ \hline (6, 0) & -6 \\ (4, 1) & -2 \\ (3, 2) & 1\end{array}\]
  4. The half-plane \(-x + 2y > 1\) does meet the region (e.g. at \((3, 5)\), \(Z = 7\)), so 1 is not the maximum: Z can be made as large as we like.
Answer: Z has no maximum value

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Exercise 12.1, Question 10

Maximise \(Z = x + y\) subject to \(x - y \le -1\), \(-x + y \le 0\), \(x, y \ge 0\).
Show solution
  1. \(x - y \le -1\) means \(y \ge x + 1\); \(-x + y \le 0\) means \(y \le x\).
  2. No point can have both \(y \ge x + 1\) and \(y \le x\): the feasible region is empty.
Answer: No feasible region, so Z has no maximum

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Textbook: NCERT Mathematics Class 12, Parts I and II (rationalised edition, 2023-24 reprint onward), free from ncert.nic.in. Question statements are shortened to the minimum needed; the solutions and tips are our own. CBSE Math Revision is independent and not affiliated with NCERT or CBSE. Spotted a slip? Tell us and it goes in the corrections log.