Maximum Cross-Sectional Area Implies Maximum Volume?

Calculus Level 3

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?

Yes. No.

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1 solution

Steven Chase
Oct 29, 2017

What ratio gives us the maximum volume? Call the horizontal dimension x x and the vertical dimension y y . The volume is (when revolved around the vertical axis):

V = π x 2 y V = \pi x^2 y

The perimeter constraint is:

P = 2 x + 2 y y = P 2 x P = 2 x + 2 y \\ y = \frac{P}{2} - x

Substituting in for y y :

V = π x 2 ( P 2 x ) = P π 2 x 2 π x 3 V = \pi x^2 \Big(\frac{P}{2} - x\Big) = \frac{P \pi}{2} x^2 - \pi x^3

Find the x 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 = P 3 P \pi x - 3 \pi x^2 = 0 \\ x = \frac{P}{3}

In the square case, x = P 4 x = \frac{P}{4} . 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.

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