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
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 $150(10)=1500$.