Maximizing the Volume of a Rain GutterDate: 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/ |
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