A toy company manufactures two types of dolls A and B. Market tests and available resources have indicated that the combined production level should not exceed 1200 dolls per week and the demand for dolls of type B is at most half of that for dolls of type A. Further, the production level of dolls of type A can exceed three times the production of dolls of other type by at most 600 units. If the company makes profit of Rs. 12 and Rs. 16 per doll respectively on dolls A and B, then how many of each should be produced weekly in order to maximise the profit?
Linear Programming — Class 12 Maths Solution
Step-by-step Solution
.: Let ‘x’ dolls of type A and ‘y’ dolls of type B be manufactured.
Then LPP problem is as below :
Maximise : $Z = 12x + 16y$ ...(1)
Subject to constraints: $x + y \le 1200$ ...(2)
$y \le \cfrac{x}{2} \Leftrightarrow x - 2y \ge 0$ ...(3)
$x \le 3y + 600 \Leftrightarrow x - 3y \le 600$ ...(4)
and $x,y \ge 0$ ...(5)
${l_1}:x + y = 1200;{l_2}:x - 2y = 0;{l_3}:x - 3y = 600$
Let us graph the inequalities (2) to (5).
The shaded portion is the feasible region which is bounded.
Let us evaluate Z at the comer points C(600, 0), F(1050, 150) and E(800, 400).
Hence, maximum profit is Rs. 16000 when 800
dolls of type A and 400 dolls of type B are manufactured and sold.
NCERT & Exemplar solution for CBSE Class 12 Mathematics, Linear Programming. Curated by Sachin Sharma. Free for all students.