aluminuim can?

Question

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 t0p and bottom pieces of height h cm are cut from square-shape aluminium sheets.
determine the value for r,h and hence calculate the ratio h/r so that the total area of the alumium sheets used for making the tin is minimum.(refer to the diagram 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 hj/r for each cases.

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 results of your investigation,what conclusion can you derive frrom the relationship of the ratio h/r and the number of sides of a regular polygon?

Answer 1

1. Here is the first part.

What is the ratio of the height to the radius of a can that has minimum surface area for a given volume?

r = radius can
h = height can
V = volume
S = surface area

V = πr²h
S = 2πr² + 2πrh

Solve for h.

V = πr²h
h = V/(πr²)

Substitute into S.

S = 2πr² + 2πrh = 2πr² + 2πr{V/(πr²)} = 2πr² + 2V/r

Take the derivative of S to find critical values.

S = 2πr² + 2V/r
dS/dr = 4πr - 2V/r² = 0
4πr = 2V/r²
2πr = V/r²
r³ = V/(2π)
r = {V/(2π)}^⅓

Take the second derivative of S to find the nature of the critical values.

dS/dr = 4πr - 2V/r²
d²S/dr² = 4π + 4V/r³ > 0 since both terms are positive
This implies a relative minimum, which is what we want.

Solve for h.

h = V/(πr²) = 2V/(2πr²) = (2/r²)(V/2π) = (2/r²)(r³) = 2r
h = 2r

Ratio of h/r

h/r = 2r/r = 2

Answer 2

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Answer 3

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Answer 4

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Answer 5

Uh...no.