Word Count : 878

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Let $x$ denote the number of type A dresses produced per day and $y$ the number of type B dresses produced per day. The manufacturer’s profit is Rs. 160 per A and Rs. 180 per B, so the objective is to maximize the total profit $$ Z \;=\; 160x \;+\; 180y. $$ The problem statement gives capacity information in terms of equivalent production: the available man-hours are enough to make either 50 pieces of A or 20 pieces of B in a day; similarly, available machine-hours suffice for either 36 pieces of A or 24 pieces of B. Translating these into linear constraints is easiest by observing that using up the entire available man-hours by producing only A means $x=50$, and producing only B means $y=20$. If both types are produced, the fraction of the man-hour resource used equals $x/50 + y/20$, which must not exceed 1. Thus the man-hour constraint is $$ \frac{x}{50} \;+\; \frac{y}{20} \;\le\; 1. $$ Similarly, for machine-hours the fraction of that resource used is $x/36 + y/24$; this must also ____ ____ ____ __________ _____ _______ _______.
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