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Two tailors P and Q earn \$150 and \$200 per day respectively. P can stitch 6 shirts and 4 trousers a day, while Q can stitch 10 shirts and 4 trousers a day. How many days should each work to produce at least 60 shirts and 32 trousers at minimum labour cost?

My solution: Let P works for x days and Q works for y days. The linear programming problem can be written as: $$ Min z= 150x+200y\\ subject \space to \ 6x+10y \geq 60 \\ 4x+4y \geq 32 \\ x\geq0 , y\geq0 $$ The grey portion is the feasible region enter image description here

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  • $\begingroup$ Is my approach correct? $\endgroup$ – user223741 Aug 29 '18 at 17:59
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Yes, your approach is correct.

I hope you obtain the point by pushing a ruler and observe when does a line of the form of $150x+200y=k$ touches the feasible region

$(5,3)$ is in fact the optimal solution attaining cost of $150(5) + 200(3)=1350$. This is lower than the other values obtained at other corners of the feasible regions which are $200(8)=1600$ or $150(10)=1500$.

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