A manufacturer makes two types of toys A and B. Three machines are needed for this purpose and the time (in minutes) required for each toy on the machines is given below:
Each machine is available for a maximum of 6 hours per day. If the profit on each toy of type A is Rs. 7.50 and that on each toy of type B is Rs. 5, then show that 15 toys of type A and 30 toys of type B should be manufactured in a day to get maximum profit.
.: Let ‘x’ toys of type A and ‘y’ toys of type B be manufactured per day.
The mathematical formulation of the problem is given below.
Maximise : $Z = \cfrac{{15}}{2}x + 5y$ …(1)
Subject to constraints : $12x + 6y \le 360 \Leftrightarrow 2x + y \le 60$ ...(2)
$18x + 0y \le 360 \Leftrightarrow x \le 20$ ...(3)
$6x9y360 \Leftrightarrow 2x + 3y \le 120$ ...(4)
$x \ge 0,y \ge 0$ ...(5)
${l_1}:2x + y = 60;{l_2}x = 20;{l_3}:2x + 3y = 120$
Let us graph the inequalities(2) to(5).
The shaded portion is the feasible region which is bounded.
Let us evaluate Z at comer points E(20, 0), F(20, 20), G(15, 30), D(0, 40).
Hence the maximum profit is Rs. 262.50
when 15 toys of type A and 30 toys of type B are manufactured.
NCERT & Exemplar solution for CBSE Class 12 Mathematics, Linear Programming. Curated by Sachin Sharma. Free for all students.