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Consider the following Linear Programming Problem :

A company produces two types of hats say A and B. Each type A hat requires twice as much labour as type B. If all the hats are of type B only, the company can produce 500 hats a day. The market limits the sale of type A and type B hats to 150 and 250 respectively. If the profit earned on type A hat is ₹ 8 and type B hat is ₹ 5, how many hats of each type should be produced so as to maximize the profit ?

The primal and dual of this problem is :

  • A.Maximize : \(Z = 8x_1 + 5x_2\) Subject to : \(2x_1 + x_2 \le 500,\ x_1 \le 150,\ x_2 \le 250,\ x_1, x_2 \ge 0\) Dual : Minimize : \(Z' = 500y_1 + 150y_2 + 250y_3\) Subject to : \(2y_1 + y_2 \le 8,\ y_1 + y_3 \le 5,\ y_1, y_2 \ge 0\)
  • B.Maximize : \(Z = 8x_1 + 5x_2\) Subject to : \(2x_1 + x_2 \le 500,\ x_1 \le 150,\ x_2 \le 250,\ x_1, x_2 \ge 0\) Dual : Minimize : \(Z' = 500y_1 + 150y_2 + 250y_3\) Subject to : \(2y_1 + y_2 \ge 8,\ y_1 + y_3 \ge 5,\ y_1, y_2 \ge 0\)
  • C.Maximize : \(Z = 8x_1 + 5x_2\) Subject to : \(2x_1 + x_2 \le 500,\ x_1 \le 150,\ x_2 \le 250,\ x_1, x_2 \ge 0\) Dual : Minimize : \(Z' = 8y_1 + 5y_2\) Subject to : \(2y_1 + y_2 \ge 500,\ y_1 \ge 150,\ y_2 \ge 250,\ y_1, y_2 \ge 0\)
  • D.Primal and dual are the same Maximize : \(Z = 8x_1 + 5x_2\) Subject to : \(2x_1 + x_2 \le 500,\ x_1 \le 150,\ x_2 \le 250,\ x_1, x_2 \ge 0\)

Answer: B

The correct pair is option B: the primal with \(\le\) constraints and the dual minimise \(500y_1 + 150y_2 + 250y_3\) subject to \(2y_1 + y_2 \ge 8\) and \(y_1 + y_3 \ge 5\).

Let \(x_1, x_2\) be the type A and type B hats. Labour: an A hat takes twice a B hat and 500 B hats fill a day, so \(2x_1 + x_2 \le 500\). Market limits give \(x_1 \le 150\) and \(x_2 \le 250\). Profit is \(Z = 8x_1 + 5x_2\).

For a maximisation primal with \(\le\) constraints, the dual is a minimisation with one variable per constraint (\(y_1, y_2, y_3\)), objective coefficients from the right-hand sides (500, 150, 250), and \(\ge\) constraints, one per primal variable, with the profits 8 and 5 on the right: column of \(x_1\) gives \(2y_1 + y_2 \ge 8\), column of \(x_2\) gives \(y_1 + y_3 \ge 5\).

Option A has the dual inequalities the wrong way round. Option C swaps the roles of costs and right-hand sides and drops a dual variable, and the primal and dual of this problem are not the same.

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