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Question:

I need help with additional mathematics project about aluminium cans.?

The Muhibbah Company is a manufacturer of cylindrical aluminium tins.The manager plans to reduce the cost of productionThe production cost is proportional to the area of aluminium sheet usedThe volume that each tin can hold is 1000 cm?( 1 litre )1.Determine the value of h, r and hence calculate the ratio of h/r when the total surface area of each tin is minimumHere, h cm denotes the height and r cm the radius of the tin.2The top and bottom pieces of the tin of height h cm are cut from square-shaped aluminium sheets.Determine the value for r,h and hence calculate the ratio h/r so that the total area of the aluminium sheets used for making the tin is minimum( refer to the diagram below )3Investigate cases where the top and bottom surfaces are cut fromi) Equilateral triangleii) Regular hexagonFind the ratio of h/r for each case.

Answer:

Talk to some pros and see if they have good used units they can install, maybe some side work, if you get a good price then he can pay for it.
There is no doubt about this problemThe former owners dog liked that unitPet urine is what caused the problemShut off the power to the unit wash it the best you can with a garden hose or power washer if you have oneThe unit will still run but its not as efficient as it once wasThe only way to repair is to replace the condensing unitcosts 1200 to 2000 good luck.
1.) The cylinder with the lowest surface area per unit volume will be cubelike Surface area for a cylinder is 2πr^2 + 2πrh, where the 2πr^2 is for the base, and 2πrh is for the tube itselfOn a similar cube, the base would be 2 sides (top and bottom), and the tube would be remaining 4 sidesThus, for a cubelike cylinder, 2πrh will equal 2 2πr^2 (since the tube accounts for twice the area as the base)2πrh 2 2πr^2 h 2r (h/r 2 for your ratio) V πr^2h 2πr^3 1000 2πr^3 500 / π r^3 159.15 r 5.419 cm, so h 10.838 cm SA 2πr^2 + 2πrh 2πr^2 + 4πr^2 6πr^2 6πr^2 553.52 cm^2 Check: r 5.418 V πr^2h 1000 h 1000 / π (5.418 ^2) 10.844 SA 2πr^2 + 2πrh 2π (5.418 ^2) + 2π (5.418 10.844) 553.60 r 5.42 V πr^2h 1000 h 1000 / π (5.42 ^2) 10.836 SA 2πr^2 + 2πrh 2π (5.42 ^2) + 2π (5.42 10.836) 553.56 Without the diagram, it'll be hard to answer 2 and 3.

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