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Given,\begin{array}{llll} & 1 & 2 & 3 \\ \text { width }(m) & 2 & 3 & 4.9 \\ \text { length } & 60 & 54 & 20 \\ \text { Standard width } & 7 & 10\end{array}let x be no. of 7 \mathrm{~m} rolls and y be no of 10 \mathrm{~m} rolls made.Now trim loss will be 7 x+10 y - total width required\begin{array}{l}=7 x+10 y-(2+3+84.9) \\=7 x+10 y-9.9\end{array}we need to minimise above such that trim loss >=0x, y>=0 and are integersTrim loss 7 x+10 y-9.9minimise trin los -9.9\begin{array}{r}(x, y)=(2,0) \Rightarrow \text { Giveo }) \text { min trim } \\\text { lors }=4.1\end{array}(x, y)=(0,1) \Rightarrow Gives hrin trim\text { loss }=0.1So, we need to manufacture 2 ex curtains of 7 \mathrm{~m} or 1 curtain of 10 \mathrm{~m} for minimum trim loss. ...