ADDITIONAL MATHEMATICS PROJECT WORK(F.4)

shada · Post time 30-10-2006 06:51 PM

ermm..nk mintak tlg ckit tunjuk ajar laa...kitorg dpt project nih dr kementerian n kena hantar be4 cuti.

ALUMINIUM CAM
The Muhibbah Company is a manufacturer of cylindrical aluminium tins. The manager plans to reduce the cost of production. The production cost is proportional to the area of the aluminium sheet used. The volume that each tin can hold is 1000cm³ (1 liter).

1)Determine the value of h, r and hence calculate the ratio of h/r when the total surface area of each tin is minimum. here h cm denotes the height and r cm the radius of the tin.

2)the 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.

3)investigate cause where the top and buttom surface are cut from
1. equllateral triangle
2.regular hexagon
find the ratio of h/r for each case.

shada · Post time 30-10-2006 07:12 PM

1 ~> vol = 1000 cm³ = pi r²h
~> h = 1000/pi r²

area ~> 2 pi r² + 2(pi)rh
      ~> 2 pi r² + 2(pi)r²[1000/pi r²]
      ~> 2 pi r² + 2000/r
dA/dr ~> 4 pi r - 2000/r = 0
      ~> 4 pi r² = 2000
      ~> pi r² = 2000/4
      ~> pi r² = 500
      ~> r² = 500/ pi
      ~> r² = 159.15
      ~> r = 12.615cm# <~ saya rasa jwpn nih x logik coz h akan = 2cm

* silap kat mana eyk?

aphrodisiac · Post time 30-10-2006 08:23 PM

1 ~> vol = 1000 cm³ = pi r²h
~> h = 1000/pi r²

area ~> 2 pi r² + 2(pi)rh
      ~> 2 pi r² + 2(pi)r[1000/pi r²]
      ~> 2 pi r² + 2000/r
dA/dr ~> 4 pi r - 2000/r = 0
      ~> 4 pi r² = 2000
      ~> pi r² = 2000/4
      ~> pi r² = 500
      ~> r² = 500/ pi
      ~> r² = 159.15
      ~> r = 12.615cm# <~ saya rasa jwpn nih x logik coz h akan = 2cm

* silap kat mana eyk?

kan sepatutnya dA/dr = 4*pi*r - 2000/r2

MACD · Post time 30-10-2006 08:56 PM

dA/dr = (4πr) - (2000/r^2 )

Utk nilai permukaan yg minimum, dA/dr = 0
(4πr) - (2000/r^2 ) = 0
(4πr) = (2000/r^2)
πr^3 = 500
r = 3√(500/π ) <---------------(punca kuasa tiga bag nilai 500/n)
r = 5.419 cm

EDIT: Simbol kuasa dua & kuasa tiga tak keluar.

[ Last edited by MACD at 30-10-2006 09:00 PM ]

shada · Post time 30-10-2006 10:04 PM

wokeh fhm. teng kiyu...

ni btol x?
2)
~> v = 4r²h = 1000cm³
~> A = 2(4r²) + h(8r)
      = 8r² + 8rh
~> h = 1000/4r²
~> dA/dr = 16r - 2000/r² = 0
                  16r = 2000/r²
      16r³ = 2000
                     r³ = 2000/16 = 125
                     r = 5 (cube root)
~> h = 1000/4r²
~> h = 1000/100
~> h = 10

Show all posts · Post time 31-10-2006 08:28 AM

Originally posted by shada at 30-10-2006 10:04 PM
wokeh fhm. teng kiyu...

ni btol x?
2)
~> v = 4r²h = 1000cm³
~> A = 2(4r²) + h(8r)
= 8r² + 8rh
~> h = 1000/4r²
~> dA/dr = 16r - 2000/r² = ...

How come dA/dr=0 ??? u just assume or ini problem pasal apa?