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Maximizing the Volume of a Rain Gutter

```Date: 10/21/2003 at 03:56:07
From: Kay
Subject: Related Rates

A rain gutter is to be constructed from a metal sheet of width 30cm,
by bending up one-third of the sheet on each side by an angle (theta)
from the horizontal (theta = zero represents the unbent sheet).
Determine what theta should be chosen so that the gutter will carry
the most water when it is full.

I figured that when theta changes, the height from the horizontal to
the top of the flap would change as well as the horizontal distance
between the flaps at the top.
```

```
Date: 10/21/2003 at 07:30:52
From: Doctor Luis
Subject: Re: Related Rates

Hello Kay,

The only variables changing in this problem are the angle theta and
the volume of water carried by the gutter (V).  As you've noted, theta
= 0 gives you zero volume (just a flat sheet), while the maximum you
can physically bend the sides is theta = 2pi/3 (giving you an
equilateral triangle cross section).

So, to find the maximum volume we need to establish a relationship
between the two variables theta and V. We can do that from the
following diagram:

The sheet has width W, length L, and the side thirds are bent at an
angle theta upwards from the horizontal.

So the volume is just equal to the length of the sheet times the cross
sectional area A:

V = L * A

Since the length is just a multiplicative constant, maximizing the
volume is the same as maximizing the cross sectional area A.

You can see from the diagrams that A consists of the area of a
trapezoid.  We know from geometry that the area of a trapezoid is
equal to half the height times the sum of the parallel sides:

A = (1/2) * h * (S1 + S2)

where

h  = (W/3)sin(theta)                (height)
S1 =  W/3                           (first parallel side)
S2 = (W/3) + 2(W/3)cos(theta)       (second parallel side)

This allows us to express the area A as a function of theta, A =
A(theta).  Keep in mind that in this specific problem we know that W =
30, so these measurements can be expressed more simply in your function.

To find the extrema points (we expect only one but who knows until you
solve the equation), solve the equation A'(theta) = 0, then check that
A"(theta) < 0 for that extremum to prove that it's a maximum.

Once you know the maximum cross section A in terms of W, you can find
the maximum volume V in terms of L and W.

As a side note, one should always check the bounds of the interval,
or any other points where the derivative might be undefined, and test
whether they are extrema.  Earlier, we showed that theta was
restricted to the interval [0,2pi/3].  While theta = 0 is certainly
an extremum, it is a minimum so we're not interested.  Theta = 2pi/3
might yield something, but that solution is physically undesirable
for the simple reason that it wouldn't make a very good gutter!  It
would be completely closed, so the volume of water that could fit
inside the triangle cross section is practically zero.  It would be
useless unless the gutter is used not to collect water, but to
transport it elsewhere, perhaps assisted by gravity.

The moral is "always check the endpoints" after you do the math.  You
might find an extrema there, although in this case the maximum value
appears inside the interval.

Lastly, here's a plot of the volume of a  W x L = 3 x 12 gutter as a
function of theta, over the interval [0,2pi/3]:

Well, I hope this helped!

Let us know if you have any more questions.

- Doctor Luis, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Calculus
High School Calculus

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