A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A require 5 minutes each for cutting and 10 minutes each for assembling. Souvenirs of type 5 require 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours 20 minutes available

Question

A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A require 5 minutes each for cutting and 10 minutes each for assembling. Souvenirs of type 5 require 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours 20 minutes available for cutting and 4 hours for assembling. The profit is Rs. 5 each for type A and Rs.6 each for type 5 souvenirs. How many souvenirs of each type should the company manufacture in order to maximise the profit ?

Accepted Answer

.: Let the company manufacture x souvenirs of type A and y souvenirs of type5.

Clearly $x \ge 0,y \ge 0.$ We make the following table from the given data.

$figure$

Since the time available for cutting is 3 hours 20 min and assembling is 4 hours,

we have the constraints
$10x + 8y \le 240$

Total profit earned is $Z = 5x + 6y$
Hence, the mathematical formulation of the problem is

Maximize $Z = 5x + 6y$ ...(1)

subject to the constraints
$5x + 8y \le 200$ ...(2)

$5x + 4y \le 120$ ...(3)

$x,y \ge 0$ ...(4)

Let ${l_1}:5x + 8y = 200$ ,

${l_2}:5x + 4y = 120$

Let us graph the inequalities (2) to (4).

$figure$

The feasible region is shown in the graph.

Here again, we observe that the feasible region is bounded.

Let us evaluate Z at the comer points
C(24, 0), E(8, 20) and B(0, 25).

We find that the maximum value of Z is 160 at B(8, 20).

Hence, the company should manufacture 8 souvenirs of

type A and 20 souvenirs of type B to realise maximum profit and maximum profit is Rs.160.

Linear Programming — Class 12 Maths Solution