This exercise is about optimization and seeks to prove that if the new strategy of packing the discs is adopted, then h/r = /n ≈2.21.
We must determine the amount of metal consumed by each end, or the area of each hexagon.
The hexagon is divided into six congruent triangles, each of which has one side (s in the diagram) in common with the hexagon.
Step I
Next, let's derive the length of s = 2r tan π/6 = (2/()r². From this we can state that the area of each of the triangles are 1/2(sr) = (1/
)r²
while the total area of the hexagon is 6 * (1/)r² = (2/
)r².
From the above, we can state that the quantity we want to minimize is given as:
A = 2πrh + 2* (2/)r²
Step 2
Next, we substitute for <em>h</em><em> </em> and differentiate. This gives us:
da/dr = - (2V/r²) + .
Let us equate the above to zero.
= 2V = 2πr²h ⇒ h/r =
/n
The above is approximately 2.21
Because d²A/dr²= + 4V/r
> 0 the above minimizes A.
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