Draw an arbitrary rectangle of the fixed perimeter on the top-right quadrant, such that two of the sides are individually collinear with the horizontal and vertical axes. We know that out of all rectangles, a square has the maximum area.
If we rotate the square about either horizontal or vertical axes, must either volumes of the right cylinder also be the maximum?
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What ratio gives us the maximum volume? Call the horizontal dimension x and the vertical dimension y . The volume is (when revolved around the vertical axis):
V = π x 2 y
The perimeter constraint is:
P = 2 x + 2 y y = 2 P − x
Substituting in for y :
V = π x 2 ( 2 P − x ) = 2 P π x 2 − π x 3
Find the x value corresponding to the max volume (the second derivative test, etc. are left as exercises for the reader):
P π x − 3 π x 2 = 0 x = 3 P
In the square case, x = 4 P . It makes sense that the volume should be maximized when the horizontal component takes up a larger share of the perimeter, since the volume depends on the square of the horizontal component, and only on the first power of the vertical component.