Please give me the answer in calculus 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.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

For a cylinder:

Volume V = π(r^2)h = 1000 cm^3
Surface Area S = 2πr^2 + 2πrh

V = π(r^2)h
h = V/(πr^2)

Solving for r and h to achieve minimum surface area:

S = 2πr^2 + 2πrh = 2πr^2 + 2πr(V/(πr^2))
S = 2πr^2 + 2V/r

dS/dr = 4πr - 2V/r^2 = 0
4πr = 2V/r^2
2πr = V/r^2
2πr^3 = V
V = 2πr^3

Use the second derivative to see if the solution is maximum or minimum surface area.

d^2S/dr^2 = 4π + 4V/r^3 > 0. So it's a minimum.

V = 2πr^3 = π(r^2)h
2r = h
h = 2r

V = 2πr^3 = 1000
πr^3 = 500
r^3 = 500/π
r = (500/π)^(1/3) =
h = 2r = 2(500/π)^(1/3)

Answer 2

Equilateral triangle or Regular hexagon....Same?

Answer 3

What is this, your final exam?