Well this is my end year exam assignment and i would appreciate it if sum1 can help me do it, I'm stuck :P.

Aluminium Can

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 Litre)

1. Determine the value ofh,rand hence calculate the ratio ofh/rwhen the total surface area of each tin is minimum. Here,hcm denotes the height andrcm the radius of the tin.

2. The top and bottom piece of the tin of heighthcm 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 cases where the top and bottom surfaces are cut from

i) Equilateral Triangle

ii) Regular hexagon

Find the ratio of h/r for each case.

Further Investigation

Investigate cases where the top and bottom faces of the tin are being cut from aluminium sheets consisting shapes of regular polygons. From the result of your investigation, what conclusion can you derive from the relationship of the ratio of h/r and the number of sides of a regular polygon?

Wastage occurs when circles are cut from aluminium sheet, which is not round in shape. Suggest the best possible shape of aluminium sheets to be used so as to reduce the production cost.

My progress so far:

I assumed that Q1 can be solved by simultaneous equation so i got these

A=Area, h=height r= radius

1000=πr²h

h=1000/πr² ---(1)

sub (1) into

A=2πrh+2πr²

A=2πr(1000/πr²+2πr²

A=(2000πr/πr²+2πr²

A=(2000r^-1)+2πr²

dA/dr=-(2000/r²+4πr

when Area is minimum, dA/dr=0

-(2000/r²+4πr=0

-2000+4πr^3=0

We have: .4πr³ .= .2000

. . . . . . . . . r³ .= .500/π

. . . . . . . . . .r .= .cbrt(500/π) .≈ .5.419

Then: .h .= .1000/πr² .= .1000/π[cbrt(500/π)]² .≈ .10.838

. . . . . . . . . . . . . . .h . . . 1000/πr² . . .1000

We have the ratio: . -- .= .---------- .= .------

. . . . . . . . . . . . . . .r . . . . . . r . . . . . . πr³

. . . . . . . . . . . . . . . . . . . . . . . .h . . . . 1000

Since r = cbrt(500/π), we have: . -- .= .----------- .= .2

. . . . . . . . . . . . . . . . . . . . . . . .r . . . π(500/π)

The height is always twice the radius.

The side view of the tin is a square.

Q1 Solved, Move on 2 Q2

Q2 must also use Simultaneous EQ

1000=πr²h

h=1000/πr² ----(1)

A=8r²+2πrh ------(2)

sub (1) into (2)

A=8r²+2πr(1000/πr²

A=8r²+(2000/r)

dA/dr=16r-2000/r²

When A minimum dA/dr=0

16r-2000/r²=0

16r³-2000=0

16r³=2000

r³=2000/16

r³=125

r=cbrt(125)

r=5

sub r= 5 into (1)

h=1000/π(5)²

h=1000/25π

h=40/π

h=12.7324

ratio h/r=(1000/πr²/r=1000/πr³

r=5, h/r=28/11

Therefore, the ratio of h:r=28:11

(Solved, please check again)Thx

Moving on 2 Q3(i)

height of equilateral triangle=ht

length of each sides of triangle=x

radius of circle=r

Ac=Area of circle

ht=x sin 60

ht=(1/2)(sqrt[3x])

r=(1/6)(sqrt[3x])

3r=(1/2)(sqrt[3x])

3r=ht

ht= x sin 60

3r/0.5=x

6r=x

.. ht=3r x=6r

Ac=3r(3r)

Ac=9r²

Total Surface Area= A

h=1000/πr²

A=18r²+2πrh

A=18r²+(2000/r)

dA/dr=36r-(2000/r²

when area min, dA/dr=0

36r-(2000/r²=0

36r³-2000=0

r³=2000/36

r³=500/9

r=3.8157

h=21.8626

h/r=(1000/πr²/r=1000/πr³

r³=500/9

h/r=1000/π(500/9)

h/r=63/11

(Q3(i) Solved)Rock on to Q3(ii)(As usual, check this please)