$($A$)$ Use the appropriate location procedure to determine the optimal coordinates of a new machine based on the following assumptions. $($Please show the details of your work; Not just the final answer$)$:Traffic pattern: RectilinearObjective: Minimizing costCoordinates of the existing machines $(5,2),(4,8),(10,12),(3,20),(7,7)$Total weight of material to be moved between the new machine and the existing ones: $11000,27500,22000,3000$ and $17000$ pounds per week, respectively.$($B$)$ Calculate cost in distance$-$weight units.Considering about the reactiliner method it's$+17000=80500$Finding new machine location in weighted average method :X cordinate :X $=(11000$ x $5)+(27500$ x$4)+(22000$ x $10)210000\mathrm{.}01\mathrm{.}147000\mathrm{.}71$lonrnna learn.bcit.caobjective is minimizing cost.MachineWeights in poundscordinates $($X$,$ Y$)111000(5,2)227500(4,8)322000(10,12)43000(3,20)517000(7,7)$Total weight $=11000+27500+22000+3000+17000=80500$Finding new machine location in weighted average method :X cordinate :X $=\{(11000$ x $5)+(27500\backslash$times $4)+(22000$ x $10)+(3000$ x $3)+(17000$ x $7)\}/80500=(55000+110000+220000+9000+119000)/80500=6\mathrm{.}37$Y cordinate :Y $=\{(11000$ x $2)+(27500$ x$8)+(22000$ x $12)+(3000$ x $20)+(17000$ x $7)3/80500=(22000+220000+264000+60000+119000)/80500=8\mathrm{.}51$cost of distance weightMachine $01$ distance $-(6\mathrm{.}37-5)+(8\mathrm{.}51-2)=7\mathrm{.}88$Machine $02$ distance $-(6\mathrm{.}37-4)+(8\mathrm{.}51-8)=2\mathrm{.}88$Machine $03$ distance $-(6\mathrm{.}37-10)+(8\mathrm{.}51-12)=7\mathrm{.}12$Machine $04$ distance $-(6\mathrm{.}37-3)+(8\mathrm{.}51-20)=14\mathrm{.}86$Machine $05$ distance $-(6\mathrm{.}37-7)+(8\mathrm{.}51-7)=2\mathrm{.}14$MAcine $01=$ DIstance x weight $=7\mathrm{.}88$ X $11000=86680$MAcine $02=2\mathrm{.}88*27500=79200$Machine $03$ distance $-(6\mathrm{.}37-10)+(8\mathrm{.}51-12)=7\mathrm{.}12$Machine $04$ distance $-(6\mathrm{.}37-3)+(8\mathrm{.}51-20)=14\mathrm{.}86$Machine $05$ distance $-(6\mathrm{.}37-7)+(8\mathrm{.}51-7)=2\mathrm{.}14$MAcine $01=$ DIstance x weight $=7\mathrm{.}88$ X $11000=86680$MAcine $02=2\mathrm{.}88$ X $27500=79200$MAcine $03=7\mathrm{.}12$ X $22000=156640$MAcine $04=14\mathrm{.}86$ X $3000=44580$MAcine $05=2\mathrm{.}14$ X $17000=36380$Total cost $=403480$