Question
A manufacturer employs three inputs: man-hours, machine-hours and cloth material to manufacture two types of dresses. Type A dress fetches him a profit of Rs. 160/- per piece, while type B, that of Rs. 180/- per piece. The manufacturer has enough man-hours to manufacture 50 pieces of type A or 20 pieces of type B dresses per day, while the machine-hours he possesses suffice only for 36 pieces of type A or for 24 pieces of type B dresses. Formulate the linear programming model and solve it graphically.
Answer :
Word Count : 815
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Let us denote: * $x$ = number of Type A dresses manufactured per day * $y$ = number of Type B dresses manufactured per day ### Objective Function: Profit from Type A dress = Rs. 160 per piece Profit from Type B dress = Rs. 180 per piece So, the total profit $Z$ is: $$ \text{Maximize } Z = 160x + 180y $$ ### Constraints: 1. Man-hours constraint: Given that the available man-hours are sufficient to manufacture 50 Type A or 20 Type B dresses per day, we convert that into a constraint by assuming the total man-hours available per day is the same for both: Let’s consider that: * Type A requires $\frac{1}{50}$ of total man-hours per piece * Type B requires $\frac{1}{20}$ of total man-hours per piece So, the total man-hours constraint is: ___ _____ __________ ______ _______ _____ _____ _________ ______ ____ _________.
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