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Question 1 :

Determine the value of h , r and then find the ratio of h/r when the total surface area of tin is minimum and h cm is high and r is radius of the base of the tin.

1. Find the value of r and h when the total surface area is minimum.

2. Volume of the tin is 1000cm³
V = 1000cm³
Π r²h = 1000cm³
h = 1000cm³/ Πr²

3. Surface area = 2 Π r² + 2 Π rh
= 2 Π r² + 2 Π r (1000 cm / Πr²)
STRATEGIES USED TO SOLVE PROBLEMS
Used two methods that are:
1. Calculus method and;
2. Trial and improvement method

CALCULUS

Area = 2 Π r² + 2 Π rh

= 2 Π r² + 2 Π r (1000 / r²)

A = 2 Π r² + 2000rˉ¹
Total surface area of each tin is minimum when dA / dR = 0

dA /dR = 4 Π r - 2000r¯²

0 = 4 Π r - 2000r¯²

4 Π r = 2000r¯²

4 Π r³ = 2000
r³ = 2000 / 4 Π

r = ³√ (2000 / 4 Π)

r = 5.419 cm

CARRYING OUT THE STRATEGIES

r = 5.419cm

h = 1000 cm² / (5.419 cm)²

= 10.840cm
Calculate the ratio h / r
h : r = 10.840 : 5.419

h = 10.840
r = 5.419
= 2.0003 cm²

h / r = 2 / 1 That is, a two to one ratio

Trial and improvement method

Table 1

r(cm)A(cm²)
12010.3914
21041.5056
3760.1893
4666.2624
5659.7850
6707.4237
7794.8929
8915.0496

From the table when the r=5 the value is minimum

Table 2
r(cm)A(cm²)
5.155.6
5.2554.5
5.3553.9
5.4553.7
5.5554.2
5.6554.9
5.7556.1
5.8557.7

Base on the table when the value of r = 5.4 cm the
surface area of the aluminium tin is minimum.

Answer Calculation

h = 1000 / pr²
= 1000 / p (5.4)²
= 58.9463

Calculate the ratio h / r

Ratio h / r = 58.9463 / 5.4

= 10.916

Thus the ratio of h / r is = 10.916

≈ 11.000

h / r = 11 / 1 That is, a eleven to one ratio

IDENTIFYING INFORMATION

Question 2 :

Aluminum plat that have rectangular shape is use to make each surface for bottom and upper of the tin with the high of the cylinder h cm

Determine the value of r, h and then calculate the h to get the total surface area of the substance r use to make the aluminum is minimum.

1. Find the total surface area that is minimum to build aluminum tin that has

2. Volume of the tin is 1000cm³
V = 1000cm³
pr²h = 1000cm³
h = 1000 / Πr²

3.Upper surface

4. Bottom surface

Area of upper and bottom surface area that is
needed is = 2( 2r x 2r )
= 8r²

Area of circular surface
= 2 pr x h
= 2 pr (1000 / pr²)

Total all surface
= 8r² +2 pr ( 1000 / pr²)

Carrying out the strategies
Calculus method
Surface area that is needed

A = 8r² + 2 pr (1000 / pr²)

A = 8r² + (2000 / r)

A = 8r² + 2000r¯¹
Total surface area of each tin is minimum when dA / dR = 0

dA / dr = 16r – (2000 / r²)

16r = 2000 / r²

16r³ = 2000

r³ = 2000 / 16

r = ³ √ (2000 / 16)

= 5 cm

Trial and improvement method

A = 8r² + 2 r (1000 / pr²) when

r = 5 cm

A = 8r² + ( 2000 / pr)

= 2008 cm²

Table 1
r (cm)Surface area(cm²)
12008
21032
3739
4628
5600
6621
7677
8762
9870
1010000

From the table when r = 5cm surface area is minimum

Table 2
r(cm)Surface area (cm²)
5.1600.237
5.2600.935
5.3602.078
5.4603.650
5.5602.636
5.6608.023
5.7610.797
5.8613.948
5.9617.463

From the table is true that when r = 5 cm the surface area is minimum

Answer

r = 5cm

h = 1000

r(cm)Surface area(cm²)
12010
21041
3760
4666
5660
6707
7795
8915
91064
101239

From table when r = 5cm surface area is minimum

Total surface area of substance needed

A = 6(3) r² + 2 pr (1000 / П r²)

A = 6(3) r² + 2000

A = 6(3) (5) ² + 2000

= 2010 cm²

Table 2

r(cm)Surface area(cm²)
4.1662.500
4.2659.511
4.3657.270
4.4655.740
4.5654.889
4.6654.684
4.7655.089
4.8656.105
4.9657.683
5.0659.807
5.1662.461
5.2665.623
5.3669.278
5.4673.410
5.5678.004
5.6683.046
5.7688.523
5.8694.424
5.9700.739

From table when r= 4.6 cm the surface area is minimum

Calculus method

Total surface area needed

A = 6(3) r² + 2 Пr (1000 / Пr²)

= 6(3) r² + (2000 / r)

Area of is minimum when dA / dR = 0
dA / dr = 12( 3) r = 2000 / r²

dA / dR = 0

12(3) r³ = 2000

r = 4.582 cm²

Answer

r = 4.582

h = 1000 / (4.582)²

= 15.161 cm
Calculate the ratio h / r

h : r = 15.161: 4.582

= 3.309 cm
≈ 3.000
h / r = 3 / 1 That is, a three to one ratio

b)Regular hexagon

Area bottom and upper surfaces that is needed

= 2 x 6 x ½ x r x 2c

= 12xr

= 12( r ) r / 3

= 12 r² / 3

Circular surfaces

Area of circular surfaces

= 2pr x h
= 2pr (1000 / pr²)
Carrying out the strategies

Trial and improvement method

Table 1
r (cm)Area surfaces (cm²)
12006
2128
3729
4610
5573
6583
7625
8693
9783
10893

Total surfaces area that needed (r = 1)

= 12 r² / 3 + 2 pr (1000 / pr²)

= 12 (1)² / 3 + 2 p(1)( 1000 / p(1)²)

= 2006cm²

Table 2
r (cm)Area surfaces
4.1604.268
4.2598.404
4.3593.219
4.4588.675
4.5584.741
4.6581.383
4.7578.576
4.8576.292
4.9574.509
5.0573.205
5.1572.359
5.2571.954
5.3571.971
5.4572.397
5.5573.215
5.6574.411
5.7575.975
5.8577.892
5.9580.154

From table when r = 5.2 cm area surfaces is minimum

Calculus method

A =12r² / 3 + 2pr (1000 / pr²)
Area is minimum, so dA / dr = 0
dA / dr = 24 r / 3 – 2002r¯²

24 r = 2002
nr(cm)h(cm)h/rA(cm²)
34.58215.1613.309655
45.00012.7322.54660
55.16311.9412.313581
65.24611.5662.25572
75.29311.3622.147567
85.32411.2302.109564
95.34411.1462.086561
105.35911.0842.068560
115.36911.0412.056558.74
125.37711.0102.048557.58
135.38410.9812.040557
145.38910.9612.034556.7
155.39310.9442.029556.3
165.39610.9302.026555.9
175.39910.9202.023555.7
185.40010.9152.021555.5
195.40310.9042.018555.3
205.40410.8992.017555.1

Base from question (1-3) I suggest aluminum plats that have hexagon shape are choose to be the top and bottom of the tin

CONCLUSION

I have found that the polygon that have many
sides have the least wastage when it is cut to form a
circle as the top and bottom of the surface of the
can. So, the production cost can be reduced.
As a decision, I suggest to Muhibbah Company
to choose the polygon material that many sides.
When the number of sides of polygon increase,
the value of h : r approaches 2.
Hence the wastage can be reduced.
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