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A-plant-group-of-company-is-to-convey-a-machinery-from-city-X-with-co-ordinate-300-200-to-city-Z-with-co-ordinate-700-900-The-machinery-can-be-air-lifted-at-a-cost-of-9-per-kilometer-or-through

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Question Number 9776 by tawakalitu last updated on 01/Jan/17
A plant group of company is to convey a machinery from city X with co ordinate (300,200) to city Z with co ordinate (700,900). The machinery can be air lifted at a cost of #9 per kilometer or through city Y with co ordinate (700,200) by an articulated truck at a cost of #7 per kilometer. Determine the route that is least expensive.
$$\mathrm{A}\:\mathrm{plant}\:\mathrm{group}\:\mathrm{of}\:\mathrm{company}\:\mathrm{is}\:\mathrm{to}\:\mathrm{convey}\:\mathrm{a}\: \\ $$$$\mathrm{machinery}\:\mathrm{from}\:\mathrm{city}\:\mathrm{X}\:\mathrm{with}\:\mathrm{co}\:\mathrm{ordinate} \\ $$$$\left(\mathrm{300},\mathrm{200}\right)\:\mathrm{to}\:\mathrm{city}\:\mathrm{Z}\:\mathrm{with}\:\mathrm{co}\:\mathrm{ordinate}\:\left(\mathrm{700},\mathrm{900}\right). \\ $$$$\mathrm{The}\:\mathrm{machinery}\:\mathrm{can}\:\mathrm{be}\:\mathrm{air}\:\mathrm{lifted}\:\mathrm{at}\:\mathrm{a}\:\mathrm{cost}\:\mathrm{of} \\ $$$$#\mathrm{9}\:\mathrm{per}\:\mathrm{kilometer}\:\mathrm{or}\:\mathrm{through}\:\mathrm{city}\:\mathrm{Y}\:\mathrm{with} \\ $$$$\mathrm{co}\:\mathrm{ordinate}\:\left(\mathrm{700},\mathrm{200}\right)\:\mathrm{by}\:\mathrm{an}\:\mathrm{articulated}\:\mathrm{truck} \\ $$$$\mathrm{at}\:\mathrm{a}\:\mathrm{cost}\:\mathrm{of}\:#\mathrm{7}\:\mathrm{per}\:\mathrm{kilometer}.\: \\ $$$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{route}\:\mathrm{that}\:\mathrm{is}\:\mathrm{least}\:\mathrm{expensive}. \\ $$
Answered by sandy_suhendra last updated on 04/Jan/17
the distance between X and Z=(√((700−300)^2 +(900−200)^2 ))=806.2 km the cost=806.2×#9=#7,255.8 the distance between X and Y=(√((700−300)^2 +(200−200)^2 ))=400 km the distance between Y and Z=(√((700−700)^2 +(900−200)^2 ))=700 km the cost=(400+700)×#7=#7,700
$$\mathrm{the}\:\mathrm{distance}\:\mathrm{between}\:\mathrm{X}\:\mathrm{and}\:\mathrm{Z}=\sqrt{\left(\mathrm{700}−\mathrm{300}\right)^{\mathrm{2}} +\left(\mathrm{900}−\mathrm{200}\right)^{\mathrm{2}} }=\mathrm{806}.\mathrm{2}\:\mathrm{km} \\ $$$$\mathrm{the}\:\mathrm{cost}=\mathrm{806}.\mathrm{2}×#\mathrm{9}=#\mathrm{7},\mathrm{255}.\mathrm{8} \\ $$$$ \\ $$$$\mathrm{the}\:\mathrm{distance}\:\mathrm{between}\:\mathrm{X}\:\mathrm{and}\:\mathrm{Y}=\sqrt{\left(\mathrm{700}−\mathrm{300}\right)^{\mathrm{2}} +\left(\mathrm{200}−\mathrm{200}\right)^{\mathrm{2}} }=\mathrm{400}\:\mathrm{km} \\ $$$$\mathrm{the}\:\mathrm{distance}\:\mathrm{between}\:\mathrm{Y}\:\mathrm{and}\:\mathrm{Z}=\sqrt{\left(\mathrm{700}−\mathrm{700}\right)^{\mathrm{2}} +\left(\mathrm{900}−\mathrm{200}\right)^{\mathrm{2}} }=\mathrm{700}\:\mathrm{km} \\ $$$$\mathrm{the}\:\mathrm{cost}=\left(\mathrm{400}+\mathrm{700}\right)×#\mathrm{7}=#\mathrm{7},\mathrm{700} \\ $$
Commented by tawakalitu last updated on 04/Jan/17
I really appreciate sir. God bless you.
$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$

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