please give me the graph and further investigation answer and also answer in trial and improvement method.?

Question

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 hold is 1000cm cube (1liter).
1. Determine the value of h, r and hence calculate the ratio of h/r when the total surface area of each tin is aluminium. 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 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.

top surface


h cmcurved surface


bottom surface

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 polygon. From the results 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.

Answer 1

(1) V = 1,000 cc = π(r^2)h
(2) A(side) = 2πrh
(3) A(t+b) = 2πr^2, BUT you want to cut the discs from square sheet, so
(4) A(t+b) = 2(2r)(2r) = 8r^2.
(5) A(total) = 2πrh + 8r^2
From (1), h = 1,000/(πr^2)
Substituting into (5)
(6) A = 2πr(1,000/(πr^2)) + 8r^2
(7) A = 2,000/r + 8r^2
For a graphical solution, set up the table below:
r. . A. . . h. . . . h/r
1 2008 318.30 318.30
2 1032 79.577 39.788
3. 738. 35.367 11.789
4. 628. 19.894 4.9735
5. 600. 12.732 2.5464
6. 621... 8.841 1.4736
The minimum radius looks to be about 5, so refining,
4.9 600.24 13.257 2.7055
5.0 600.00 12.732 2.5464
5.1 600.23 12.237 2.3996
Refining again,
4.99 600.002 12.783 2.5618
5.00 600.000 12.732 2.5464
5.01 600.002 12.681 2.5312

A(top) = 4*5^2 = 100 cm^2 (21.46 cm^2 waste)

A(tube) = 2π*5*12.732 = 399.98 cm^2 ≈ 400 cm^2

A(bottom) = 100 cm^2

3.
i. For equilateral triangles
A(t) becomes r²3√3 = 6r²√3
(8) A(t+b) =6r²√3 (waste is 4.1r²)
Repeat the process of finding minimum total area.
ii For regular hexagons
A(t) = 6r²/√3 = 2r²√3
(9) A(t+b) = 4r²√3 (waste = 0.645r^2)
Repeat the process of finding minimum total area.

Note that differentiating the equation for total area and setting the derivative = 0 will yield the optimum r as a solution of one equation.

Investigation of other regular polygons is fruitless, since only the triangle, square, and hexagon cover a large sheet without gaps. Of these, you waste 1 triangle per row and 1 hexagon every other row.

Answer 2

If that weren't such a freaking long problem maybe i would do your homework for you. Or, maybe you should do your own homework, otherwise you'll probably fail the class. Do you really think anyone is going to do that whole thing for you? haha.

Answer 3

are you asking us to do your homework? if yes, try to make the question simple. it's too long...