maximum volume?

Question

a rectangle of fixed perimeter 36 is rotated around one of its sides, thus sweeping out a figure in the shape of a circular cylinder. what is the maximum possible volume of that cylinder?
the length of 2 sides of the rectangle is 'h' and is also the height of the cylinder.
the length of the other 2 sides is 'r' and also the lenght of the radius of the circular cylinder.

Answer 1

1. The perimeter is 2(r + h)=36

2r + 2h = 36 // divide by 2

r + h = 18

or h=18 - r

2. The volume is v=pi*r^2 * h = pi*r^2 * (18 - r)

To find the maximum let's find where dv/dr=0

dv/dr = pi * 2r(18 - r) - pi*r^2 =
= 2pi * 18r - 2pi*r^2 - pi*r^2 = 36pi * r - 3pi*r^2

36pi * r - 3pi*r^2 = 0 // Cool, we can divide by pi !

36r - 3r^2 = 0

3r(12 - r) = 0

If r=0, the volume is 0 too.
Let's take r=12

Minimum or maximum?

Let's derive dv/dr

d^2 v / dr^2 = 36pi - 6pi*r = 36pi - 3*24*pi =
= 36pi - 72pi = -36pi < 0 -> Maximum

Now, the maximum volume is when r=12

v=pi*12^2 * (18 - 12) = pi * 144 * 6 =
= 864 pi = 2714.336...

Answer 2

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let h be the height of the resulting cylinder.

The radius of the cylinder is 18 - h = r
V = πr²h

V(h) = π (18-h)² h
V(h) = π (324h - 36h² + h³)

The extremum occurs at V' = 0
V' = 0 yields 324 - 72h + 3h² = 0
h² - 24h + 108
(h-18)(h-6) = 0

then h = 6 and r = 12. (it is a maximum by 2nd derivative test, not shown here.)
V = 864π

Answer 3

volume = pi r^2 h

perimeter = 2h + 2(2 pi r ) = 36
2 sides of rectangle is h, also the height of the cylinder
the other 2 sides is (2pi r) in relation to the radius of cylinder
---------> h = [36 - (4 pi r)] / 2
---------> h = 18 - 2pi r
volume = pi r^2 (18 - 2pi r)
------------>= 18 pi r^2 - 2pi^2 r^3

differentiate with respect to r to find maximum volume:
dV/dr = 36 pi r - 6 pi^2 r^2 = 0
36 pi r = 6 pi^2 r^2
6 = pi r
r = 6/pi

h = 18 - pi r
h = 18 - 6
h = 12

volume therefore is: pi r^2 h
------------------------> pi (6/pi)^2 (12)
------------------------> 432/pi or 137.51