Post by Syah on Jan 6, 2007 8:35:17 GMT -5
Soalan no 3 tu translate sendiri
Total surface area of each tin,A = 2h:Ðr + 2¦Ðr2
Volume of each tin = 1000 cm3
h¦Ðr2 = 1000
Therefore, h = 1000/(¦Ðr2)
A = 1000/(¦Ðr2) x 2¦Ðr + 2¦Ðr2
= 2000/r + 2¦Ðr2
dA/dr = -2000/r2 + 4¦Ðr
For minimum value of A, dA/dr = 0
Hence, 4¦Ðr = 2000/r2
r3 = 500/¦Ð
r = 5.42 cm (2 dec pl)
Since previously we say h¦Ðr2 = 1000,
h = 1000/(¦Ðr2)
= 10.84 cm (2 dec pl)
Hence the ratio h/r when area is minimum will be:
h/r = 2
2)For total area of tha aluminium sheets used for making the tin to be minimum,
Area of sheet used = 2h¦Ðr + 2 x 4r2
Similar to answer in (1), h = 1000/¦Ðr2
Hence, A = 2000/r + 8r2
dA/dr = -2000/r2 + 16r
For minimum value of A,
dA/dr = 0
16r = 2000/r2
r3 = 125
r = 5 cm
Therefore, h = 1000/¦Ðr2
= 12.73 cm (2 dec pl)
And h/r when A is minimum:
h/r = 2.5
3)
tan 60˚= x
r
Therefore , x = 1.73r
Surface area ∆PQR = 6 ¡Ñ 1 ¡Ñ x ¡Ñ r
2
= 3 ( 1.73r )r
= 5.19 r2
Total surface area of aluminium sheet
= 2 ( 5.19 r2 ) + 2£krh
= 10.38r2 + 2£kr ( 1000 )
£kr2
= 10.38r2 + 2000
r2
dA = 20.76r ¡V 2000
dr r2
For minimum area , dA = 0
dr
Therefore , 20.76r = 2000 = 0
r2
20.76r = 2000
r2
r3 = 2000
20.76
= 96.3391
r = 4.5842
Substitute r into ①
Therefore ,h= 1000___
£k(4.5842)2
= 15.1746
h=15.1746
r 4.5842
= 3.3101
SOLUTION 2
1) A can with constant volume could be tall and narrow, or short and wide, and both of these would have a lot of area. The one with the least area is in between these two, and has a nearly square profile; meaning h ~ 2r (but not exactly it turns out).
2) You can imagine that if you plotted area vs. h or r, the curve would start high, and as h or r increased, it would slope down to a minimum area, then go back up again. You know you're at the minimum area when the slope of that curve is zero.
Assume the volume is constant = 1000 cm^3
(for consistancy use units of cm)
Volume of a cylinder = area of the circular base times the height
V = pi * r^2 * h = 1000 cm^3
Area = area of top + bottom + side
A = 2 (pi r^2) + 2 pi r h
You have 2 equations (for A and V), and 2 unknowns (r and h).
solve this in the usual way, by solving one equation for one variable (the easiest one) and substituting into the other equation.
Using:
V = pi * r^2 * h = 1000 cm^3
h = 1000 / (pi * r^2)
3) Now substitute this eqn for h into the eqn for A, and you get:
A = 2 pi r^2 + 2000/r
now take the derivative w.r.t. r
dA / dr = 4 pi r - 2000 / r^2
set this equal to zero, and solve for r.
r = cube root of (500/pi) ~= 5.419
solve for h = 10.8
check that h ~= 2r
h/r ~= 1/2
4) 2 & 3 ) Use the same method, but adjust the equation for the Area so that instead of the area of the can, the eqn is for the original stock material that the can is made of.
* V, volume for cylinder = pi r² h = 1000 cm³
A, total surface area for cylinder = 2 (pi r² h) + 2 pi r h
Since there’s 2 variables in the formulae, get rid one of them, either r or h. Here, I will get rid of r:
pi r² h = 1000
h = 1000/(pi r²)
When h = 1000/(pi r² h), A = 2000 + 2000/r
So, count dA/dr = 0, you should be able to get the h/r ratio at minimum.
SOLUTION 3
Kaji kes yang permukaan atas dan bawah tin dipotong daripada kepingan-kepingan aluminium yang berbentuk;
i)segi tiga sama sisi
ii)heksagon sekata
Kira nilai h/r untuk setiap kes.
Kesimpulan:
Lebih banyak sisi poligon sekata yang dibentuk daripadanya, lebih rendah nilai h/r tetapi nilai minimum adalah h/r=2 iaitu berbentuk bulatan
3)i)segi tiga sama sisi
t = Tinggi segitiga sama kaki
t = r + r/sin 30
t = r + 2r
t = 3r
y = panjang sisi segitiga
y = t/sin 60
y = 3r/sin 60
y = 6r/sqrt(3)
Luas satu segitiga = yt/2 = 9r^2/sqrt(3)
h = 1000/(pi.r^2)
A = 2pi.r x h + [2 x 9r^2/sqrt(3)]
A = 2pi.r x 1000/(pi.r^2) + [2 x 9r^2/sqrt(3)]
A = 2000/r + 18r^2/sqrt(3)
dA/dr = -2000/r^2 + 36r/sqrt(3)
Untuk A minimum, dA/dr=0,
-2000/r^2 + 36r/sqrt(3) = 0
36r/sqrt(3) = 2000/r^2
r^3 = 2000.sqrt(3)/36
r = 4.5824321
h = 1000/[pi(4.5824321)^2]
h = 15.158563
h/r = 3.308
3)ii)heksagon sekata
y = panjang sisi heksagon sekata
y = 2 x r/tan 60
y = 2 x r/sqrt(3)
y = 2r/sqrt(3)
Luas satu heksagon = 6 x [r.2r/sqrt(3)]/2
= 6r^2/sqrt(3)
h = 1000/(pi.r^2)
A = 2pi.r x h + [2 x 6r^2/sqrt(3)]
A = 2pi.r x 1000/(pi.r^2) + 12r^2/sqrt(3)
A = 2000/r + 12r^2/sqrt(3)
dA/dr = -2000/r^2 + 24r/sqrt(3)
Untuk A minimum, dA/dr=0,
-2000/r^2 + 24r/sqrt(3) = 0
24r/sqrt(3) = 2000/r^2
r^3 = 2000.sqrt(3)/24
r = 5.2455753
h = 1000/[pi(5.2455753)^2]
h = 11.568149
h/r = 2.205
2) Kepingan-kepingan aluminium yang berbentuk segi empat sama digunakan untuk membuat setiap permukaan atas dan bawah tin dengan tinggi silinder h cm.
h = 1000/(pi.r^2)
A = 2pi.r x h + 2(2r)^2
A = 2pi.r x 1000/(pi.r^2) + 8r^2
A = 2000/r + 8r^2
dA/dr = -2000/r^2 + 16r
Untuk A minimum, dA/dr=0,
-2000/r^2 + 16r = 0
16r = 2000/r^2
r^3 = 2000/16
r = 5cm
h = 1000/(pi.r^2)
h = 1000/(pi.5^2)
h = 12.732395
h/r = 2.546
1)
A = luas kepingan
Diberi isipadu:
pi x r^2 x h = 1000
h = 1000/(pi.r^2)
Jumlah luas digunakan:
2pi.r^2 + 2pi.rh = A
2pi.r^2 + 2pi.r[1000/(pi.r^2)] = A
A = 2pi.r^2 + 2000/r
dA/dr = 4pi.r - 2000/r^2
Untuk luas minimum (since coefficient of r^2 is positive), dA/dr=0
4pi.r - 2000/r^2 = 0
r^3 = 500/pi
r = 5.419261cm
h = 1000/(pi.r^2)
h = 1000/[pi(5.419261)^2]
h = 10.8385214cm
h/r = 2
Total surface area of each tin,A = 2h:Ðr + 2¦Ðr2
Volume of each tin = 1000 cm3
h¦Ðr2 = 1000
Therefore, h = 1000/(¦Ðr2)
A = 1000/(¦Ðr2) x 2¦Ðr + 2¦Ðr2
= 2000/r + 2¦Ðr2
dA/dr = -2000/r2 + 4¦Ðr
For minimum value of A, dA/dr = 0
Hence, 4¦Ðr = 2000/r2
r3 = 500/¦Ð
r = 5.42 cm (2 dec pl)
Since previously we say h¦Ðr2 = 1000,
h = 1000/(¦Ðr2)
= 10.84 cm (2 dec pl)
Hence the ratio h/r when area is minimum will be:
h/r = 2
2)For total area of tha aluminium sheets used for making the tin to be minimum,
Area of sheet used = 2h¦Ðr + 2 x 4r2
Similar to answer in (1), h = 1000/¦Ðr2
Hence, A = 2000/r + 8r2
dA/dr = -2000/r2 + 16r
For minimum value of A,
dA/dr = 0
16r = 2000/r2
r3 = 125
r = 5 cm
Therefore, h = 1000/¦Ðr2
= 12.73 cm (2 dec pl)
And h/r when A is minimum:
h/r = 2.5
3)
tan 60˚= x
r
Therefore , x = 1.73r
Surface area ∆PQR = 6 ¡Ñ 1 ¡Ñ x ¡Ñ r
2
= 3 ( 1.73r )r
= 5.19 r2
Total surface area of aluminium sheet
= 2 ( 5.19 r2 ) + 2£krh
= 10.38r2 + 2£kr ( 1000 )
£kr2
= 10.38r2 + 2000
r2
dA = 20.76r ¡V 2000
dr r2
For minimum area , dA = 0
dr
Therefore , 20.76r = 2000 = 0
r2
20.76r = 2000
r2
r3 = 2000
20.76
= 96.3391
r = 4.5842
Substitute r into ①
Therefore ,h= 1000___
£k(4.5842)2
= 15.1746
h=15.1746
r 4.5842
= 3.3101
SOLUTION 2
1) A can with constant volume could be tall and narrow, or short and wide, and both of these would have a lot of area. The one with the least area is in between these two, and has a nearly square profile; meaning h ~ 2r (but not exactly it turns out).
2) You can imagine that if you plotted area vs. h or r, the curve would start high, and as h or r increased, it would slope down to a minimum area, then go back up again. You know you're at the minimum area when the slope of that curve is zero.
Assume the volume is constant = 1000 cm^3
(for consistancy use units of cm)
Volume of a cylinder = area of the circular base times the height
V = pi * r^2 * h = 1000 cm^3
Area = area of top + bottom + side
A = 2 (pi r^2) + 2 pi r h
You have 2 equations (for A and V), and 2 unknowns (r and h).
solve this in the usual way, by solving one equation for one variable (the easiest one) and substituting into the other equation.
Using:
V = pi * r^2 * h = 1000 cm^3
h = 1000 / (pi * r^2)
3) Now substitute this eqn for h into the eqn for A, and you get:
A = 2 pi r^2 + 2000/r
now take the derivative w.r.t. r
dA / dr = 4 pi r - 2000 / r^2
set this equal to zero, and solve for r.
r = cube root of (500/pi) ~= 5.419
solve for h = 10.8
check that h ~= 2r
h/r ~= 1/2
4) 2 & 3 ) Use the same method, but adjust the equation for the Area so that instead of the area of the can, the eqn is for the original stock material that the can is made of.
* V, volume for cylinder = pi r² h = 1000 cm³
A, total surface area for cylinder = 2 (pi r² h) + 2 pi r h
Since there’s 2 variables in the formulae, get rid one of them, either r or h. Here, I will get rid of r:
pi r² h = 1000
h = 1000/(pi r²)
When h = 1000/(pi r² h), A = 2000 + 2000/r
So, count dA/dr = 0, you should be able to get the h/r ratio at minimum.
SOLUTION 3
Kaji kes yang permukaan atas dan bawah tin dipotong daripada kepingan-kepingan aluminium yang berbentuk;
i)segi tiga sama sisi
ii)heksagon sekata
Kira nilai h/r untuk setiap kes.
Kesimpulan:
Lebih banyak sisi poligon sekata yang dibentuk daripadanya, lebih rendah nilai h/r tetapi nilai minimum adalah h/r=2 iaitu berbentuk bulatan
3)i)segi tiga sama sisi
t = Tinggi segitiga sama kaki
t = r + r/sin 30
t = r + 2r
t = 3r
y = panjang sisi segitiga
y = t/sin 60
y = 3r/sin 60
y = 6r/sqrt(3)
Luas satu segitiga = yt/2 = 9r^2/sqrt(3)
h = 1000/(pi.r^2)
A = 2pi.r x h + [2 x 9r^2/sqrt(3)]
A = 2pi.r x 1000/(pi.r^2) + [2 x 9r^2/sqrt(3)]
A = 2000/r + 18r^2/sqrt(3)
dA/dr = -2000/r^2 + 36r/sqrt(3)
Untuk A minimum, dA/dr=0,
-2000/r^2 + 36r/sqrt(3) = 0
36r/sqrt(3) = 2000/r^2
r^3 = 2000.sqrt(3)/36
r = 4.5824321
h = 1000/[pi(4.5824321)^2]
h = 15.158563
h/r = 3.308
3)ii)heksagon sekata
y = panjang sisi heksagon sekata
y = 2 x r/tan 60
y = 2 x r/sqrt(3)
y = 2r/sqrt(3)
Luas satu heksagon = 6 x [r.2r/sqrt(3)]/2
= 6r^2/sqrt(3)
h = 1000/(pi.r^2)
A = 2pi.r x h + [2 x 6r^2/sqrt(3)]
A = 2pi.r x 1000/(pi.r^2) + 12r^2/sqrt(3)
A = 2000/r + 12r^2/sqrt(3)
dA/dr = -2000/r^2 + 24r/sqrt(3)
Untuk A minimum, dA/dr=0,
-2000/r^2 + 24r/sqrt(3) = 0
24r/sqrt(3) = 2000/r^2
r^3 = 2000.sqrt(3)/24
r = 5.2455753
h = 1000/[pi(5.2455753)^2]
h = 11.568149
h/r = 2.205
2) Kepingan-kepingan aluminium yang berbentuk segi empat sama digunakan untuk membuat setiap permukaan atas dan bawah tin dengan tinggi silinder h cm.
h = 1000/(pi.r^2)
A = 2pi.r x h + 2(2r)^2
A = 2pi.r x 1000/(pi.r^2) + 8r^2
A = 2000/r + 8r^2
dA/dr = -2000/r^2 + 16r
Untuk A minimum, dA/dr=0,
-2000/r^2 + 16r = 0
16r = 2000/r^2
r^3 = 2000/16
r = 5cm
h = 1000/(pi.r^2)
h = 1000/(pi.5^2)
h = 12.732395
h/r = 2.546
1)
A = luas kepingan
Diberi isipadu:
pi x r^2 x h = 1000
h = 1000/(pi.r^2)
Jumlah luas digunakan:
2pi.r^2 + 2pi.rh = A
2pi.r^2 + 2pi.r[1000/(pi.r^2)] = A
A = 2pi.r^2 + 2000/r
dA/dr = 4pi.r - 2000/r^2
Untuk luas minimum (since coefficient of r^2 is positive), dA/dr=0
4pi.r - 2000/r^2 = 0
r^3 = 500/pi
r = 5.419261cm
h = 1000/(pi.r^2)
h = 1000/[pi(5.419261)^2]
h = 10.8385214cm
h/r = 2
