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# 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.Formulate this problem at a LPP to maximise the profit to the company.

$\begin{array}{1 1}(A)\;z=120x+200y \\ (B)\;z=360x+120y \\ (C)\;z=200x+120y \\(D)\;z=120x+360y \end{array}$

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:
For type A: No of sweaters=360.Cost per sweater is 200.
For type B: No of sweaters=100.Cost per sweater is 120.
Maximum cost =72000.
Step 2:
Let $x$ be number of sweaters of type A and $y$ be the number of sweaters of type B.
These sweaters are brought to fulfill the maximum requirement of $x,y$ and maximize the profit.
Step 3:
The mathematic formulation of the above problem is as follows :
Maximize $z=200x+120y$
Subject to $360x+120y\leq 72,000$
$\Rightarrow 3x+y\leq 600$
$x+y\leq 300$
$y-x \leq 100$ and $x\geq 0$ and $y \geq 0$