Comment
Share
Q)

# A company manufactures two types of sweaters :type A sweaters type B.It costs Rs 360 to make a type A sweater and Rs 120 to make a type B sweater.The company can make at most 300 sweaters and spend at most Rs72,000 a day.The number of sweaters of type B cannot exceed the number of sweaters of type A by more than 100.The company makes a profit of Rs 200 for each sweater of type A and Rs 120 for every sweater of type B.What is the maximum profit (in Rs.)?

Comment
A)
Toolbox:
• Let $R$ be the feasible region for a linear programming problem and let $z=ax+by$ be the objective function.When $z$ has an optimum value (maximum or minimum),where variables $x$ and $y$ are subject to constraints described by linear inequalities,this optimum value must occur at a corner point of the feasible region.
• If R is bounded then the objective function Z has both a maximum and minimum value on R and each of these occur at corner points of R
Step 1:
Let $x$ be the no of sweaters of type A and y be the no of sweaters of type B.
It is given that it costs Rs 360 to make type A and Rs120 to make type B.
It can spend at most of Rs72000 per day.
$\therefore 360x+120y\leq 72000$
$\Rightarrow 3x+y\leq 600$
It is given that the company can product at most of 300 sweaters of both type per day.
$\therefore x+y\leq 300$
It is also given that type B no of sweater cannot exceed no of type A sweater by more than 100.
$\therefore y-x\leq 100$
The company makes a profit of Rs200 of type A sweaters and Rs 120y for type B sweater.
Hence to maximize the profit,the objective function $Z=200x+120y$
Step 2:
Now let us draw the graph for the lines
$AB :3x+y=600,CD :x+y=300$ and $EF :y-x=100$
Consider the line $AB :3x+y=600$
Put $x=0,y=0$ then $0\leq 600$ is true.
Hence the region $3x+y\leq 600$ lies below the line AB.
Consider the line $CD :x+y=300$
Put $x=0,y=0$ then $0\leq 300$ is true.
Hence the region $x+y\leq 300$ lies below the line CD.
Consider the line $EF :y-x=100$
Put $x=0,y=0$ then $0\leq 100$ is true.
Hence the region $y-x\leq 100$ lies below the line EF.
The feasible region OXPQB is the shaded portion shown in the fig.
The coordinates of $P$ is (100,200)
The coordinates of $Q$ is (150,150)
The corner points are $O(0,0),X(0,100),P(100,200),Q(150,150),B(200,0)$
Step 3:
Let us obtain the values of the objective function $Z=200x+120y$
At the points $(x,y)$ the value of the objective function subjected to $Z=200x+120y$
At $O(0,0)$ the value of the objective function $Z=0$
At $X(0,100)$ the value of the objective function $Z=200\times 0+120\times 100=12000$
At $P(100,200)$ the value of the objective function $Z=200\times 100+120\times 200=20000+24000=44000$
At $Q(150,150)$ the value of the objective function $Z=200\times 150+120\times 150=30000+18000=48000$
At $B(300,0)$ the value of the objective function $Z=200\times 300+0=60000$
The maximum profit is at $Rs.48000$ at $Q(150,150)$
Hence 15 sweaters of each sweater should be produced for a maximum profit of Rs.48,000