How to solve the Additional Mathematics Project Work Form 4 Year 2003?

Question

The Muhibah 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 cm63 ( 1 liter).
1. Determne 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 sqyare-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 the 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.

Answer 1

Area A = area of two ends + Curved surfaces.
A = 2π r² + 2π r h

Volume V = Area of end x Heigh.
V = π r² * h
V = π r² h
h = V / π r²
Now substitute for h into the area formula.

A = 2π r² + 2π r h
A = 2π r² + 2π r (V / π r²)
A = 2π r² + 2V / r
The volume of the can is 1000cm^3, so substitue in.
A = 2π r² + 2πV / r
A = 2π r² + 2000 / r

Take first derivide and let it equat to zero (zero for max. or min. pt.).
A = 2π r² + 2000 / r
δA/δr = 4π r - 2000 / r²
4π r - 2000 / r² = 0
4π r = 2000 / r²
4π r^3 = 2000
r^3 = 2000/ 4π
r = (2000/ 4π)^1/3 (that is, cubed root).
r = 5∙419 260 701
r ≈ 5∙42 cm.

Let's use the r value to calculate h.
V = π r² * h
1000 = π (5∙419 260 701)² * h
h = 1000 /π (5∙419 260 701)²
h = 10∙838 5214
h ≈ 10 51 cm.

For ratio:
h/r = 10∙838 5214 / 5∙419 260 701
h/r = 2/1 That is, a two to one ratio.

For queation two, follow the same proceedure after you set up your formulas.
Like wise for question three.

With question two it's not clear what value you are taken for the length of the side of the square edge. Because you are still using r, I'm assuming you are taken the largest size of a square that fits into the circle or radius r. Based on this assumsion, the calculations are are follows:

To identify the length x of one side of the square of radius r:
Draw two diameters of the circle at 90º to each other. Where these diameters make contact with the circumference of the circle will be the corner points of the square. The area of this square now consists of four triangles. These triangles will have two sides of length r with the enclosed angle of 90º. So the length of x will be:

x² = r² + r²
x² = 2 r²
x = √(2 r²) That is, length of one side of the square.

Area A = area of two square ends + parameter of square * heigh.
A = (2r² * 2) + [(√2)r * 4 * h]
A = 4r² + 4√(2)r h

Volume V = square area x heigh.
V = 2 r² * h
V = 2 r² h
h = V / 2 r²
Now substitute for h into the area formula.

A = 4r² + 4√(2)r h
A = 4r² + 4√(2)r V / 2 r²
A = 4r² + 2√(2) V/r
A = 4r² + 2000√(2) r^-1 (Volume is 1000cm^3).

Take first derivide and let it equat to zero (zero for max. or min. pt.).
A = 4r² + 2000√(2) r^-1
δA/δr = 8r - 2000√(2) r^-2
8r - 2000√(2) r^-2 = 0
8r = 2000√(2) r^-2
8r = [2000√(2)] /r²
8r^3 = 2000√(2)
r^3 = [2000√(2)]/ 8
r = ([2000√(2)]/ 8)^1/3 (that is, cubed root).
r = (353∙55 33 906...)^1/3
r = 7∙071 0678 12
r ≈ 7∙07 cm.

Let's use the r value to calculate h.
V = 2r² * h
1000 = 2r² * h
1000 = 2 (50) * h
1000 = 100h
h = 10 cm.

For ratio:
h/r = 10 / 7∙071 0678 12
h/r = 1∙845 270 149 / 1
h : r ≈ 1∙85 : 1
These ratios do not take into account the scrap metal from the circle when the square is cut out of it.

Answer 2

Since you only asked "how to solve" this, I'll just give you an outline of how to solve the problems.

First, you need to express the area (A) of the cross section of the tin (a circle for the first problem, square for the second, etc) in terms of r. Note that r as a radius only really makes sense for the first problem. For the other problems, r needs to represent something else, maybe half the length of a side of the square in the second problem, for example.

Then you need to express the surface area (S) of the tin in terms of r and h.
Since the goal is to minimize surface area, we need to differentiate surface area (S) with respect to r, and set that to zero.

Now you have two simultaneous equations in r and h.

Ah = 1.0
S' = 0

Solve these two equations to get your answer.
Good Luck :-)

Answer 3

The surface area of a cylinder is

A = 2 * pi * r * h

plus the area of the top and bottom

A = 2 * pi * r * h + 2 * pi * r^2

The volume of the cylinder is

V = pi * r^2 * h = 1
h = 1 / (pi * r^2)

Substituting,

A = [2 * pi * r / (pi * r^2) ] + 2 * pi * r^2