# Maths assignment

• Oct 19th 2006, 01:04 AM
bryankek
Maths assignment
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 1000cm&#179;(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 minimum. Here, h cm denotes the height and r cm the radius of the tin.

2. The top and bottom piece 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 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

1000=πr&#178;h
h=1000/πr&#178; ---(1)

sub (1) into

A=2πrh+2πr&#178;
A=2πr(1000/πr&#178;)+2πr&#178;
A=(2000πr/πr&#178;)+2πr&#178;
A=(2000r^-1)+2πr&#178;

dA/dr=-(2000/r&#178;)+4πr

when Area is minimum, dA/dr=0

-(2000/r&#178;)+4πr=0
-2000+4πr^3=0

We have: .4πr&#179; .= .2000
. . . . . . . . . r&#179; .= .500/π
. . . . . . . . . .r .= .cbrt(500/π) .≈ .5.419

Then: .h .= .1000/πr&#178; .= .1000/π[cbrt(500/π)]&#178; .≈ .10.838

. . . . . . . . . . . . . . .h . . . 1000/πr&#178; . . .1000
We have the ratio: . -- .= .---------- .= .------
. . . . . . . . . . . . . . .r . . . . . . r . . . . . . πr&#179;

. . . . . . . . . . . . . . . . . . . . . . . .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=πr&#178;h
h=1000/πr&#178; ----(1)

A=8r&#178;+2πrh ------(2)

sub (1) into (2)

A=8r&#178;+2πr(1000/πr&#178;)
A=8r&#178;+(2000/r)

dA/dr=16r-2000/r&#178;

When A minimum dA/dr=0

16r-2000/r&#178;=0
16r&#179;-2000=0
16r&#179;=2000
r&#179;=2000/16
r&#179;=125
r=cbrt(125)
r=5

sub r= 5 into (1)

h=1000/π(5)&#178;
h=1000/25π
h=40/π
h=12.7324

ratio h/r=(1000/πr&#178;)/r=1000/πr&#179;

r=5, h/r=28/11

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

Moving on 2 Q3(i)

height of equilateral triangle=ht
length of each sides of triangle=x
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=9r&#178;

Total Surface Area= A
h=1000/πr&#178;
A=18r&#178;+2πrh
A=18r&#178;+(2000/r)
dA/dr=36r-(2000/r&#178;)

when area min, dA/dr=0
36r-(2000/r&#178;)=0
36r&#179;-2000=0
r&#179;=2000/36
r&#179;=500/9
r=3.8157

h=21.8626

h/r=(1000/πr&#178;)/r=1000/πr&#179;

r&#179;=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)
• Oct 19th 2006, 06:25 AM
Soroban
Hello, Bryankek!

Your work on #1 is correct . . . excellent work!
. . I didn't approximate like you did.

Quote:

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 1000 cm³(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 minimum.
Here, h denotes the height and r the radius of the tin.

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 .= .0

Multiply by r²: . -2000 + 4πr³ .= .0
. . 4πr³ .= .2000
. . . . .= .1750/11 . . . ok

We have: .4πr³ .= .2000
. . . . . . . . . .= .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.

• Oct 20th 2006, 06:09 AM
bryankek
wow!thx, am working on 2nd 1 now