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# A small firm manufactures items A and B. The total number of items that it can manufacture in a day is at the most 24 items. A takes one hour to make while item B takes only half an hour. The maximum time available per dat is 16 hours. If the profit on one unit of item A be Rs. 300 and that of one unit of item B be Rs. 160, how many of each type of item should be produced to maximize the profit? Solve the problem graphically.

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• 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 the no of items manufactured by the firm of type A be x
Let the no of items manufactured by the firm of type B be y
It is said that the firm can manufacture items of both type is at most 24
$\Rightarrow x+y\leq 24$
The maximum time available for both A and B is 16.
Where A takes 1 hour and B takes $\large\frac{1}{2}$ an hour.
$\therefore x+\large\frac{1}{2}$$y\leq 16 The profit of one unit of item A is Rs.300 and that of B is Rs160 Maximum profit to be attained is given by Z=300x+160y Clearly x,y\geq 0 Step 2: Now let us solve the above problem graphically : Let us draw the lines x+y=24 and x+\large\frac{1}{2}$$y=16$ on the graph.
Clearly the shaded portion is the feasible region.
The corner points are OACD where coordinates are $O(0,0),A(16,0),C(8,16),D(0,24)$
Step 3:
Let us obtain the value of objective function as follows :
At the points $(x,y)$ the value of the objective function subjected to $Z=300x+160y$
At $O(0,0)$ the value of the objective function $Z=0$
At $A(16,0)$ the value of the objective function $Z=300\times 8+160\times 0=4800$
At $C(8,16)$ the value of the objective function $Z=300\times 8+160\times 16=4960$
At $D(0,24)$ the value of the objective function $Z=300\times 0+160\times 24=3840$
Step 4:
Clearly the maximum value is $Z=4960$ at $C(8,16)$
Hence 8items of type A and 16 items of type B are required to acquire a maximum profit.
edited Sep 30, 2013