Optimizing Gutter Area Essay

1086 Words Nov 1st, 2007 5 Pages
Welbilt Inc.
7439 Allen Highway
Vida, MT 59274

Dear Mr. Cal Q. Luss:

As we were approaching the problem of how to design a gutter which would carry the most water from the roof with a 12 inch wide piece of material, we came up with a thesis; the length of the roof can be any number according to the roof sections, the maximum volume of the gutter will then depend on the maximum area out of the shapes' side (flank) with respect to the 12in-wide-material piece. Thus the areas of four different shapes were found. By process of elimination, the optimal area for each shape was found by deriving the equations of area as well as basic geometry. The shapes used for this project were a semi-circle, triangle, square, and rectangle. For
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We refer to these angles as θ, and the base as b. Since the height creates two equal triangles, BD = BC = b/2. If we just look at triangle ADC alone, we can use trigonometry to find out what θ is. sinθ = AD/AC => sinθ = h/6 => h = 6sinθ (1) cosθ = DC/AC => cosθ = (b/2)/6 => cosθ = b/12 => b = 12 cosθ (2)
The equation for the area of the triangle is A = 1/2bh, we can then plug in (1) and (2) in the area equation to make it all in θ form.
A = ½ (6sinθ)(12cosθ) = 36sinθcosθ (3)
We now have to take the derivative of the simplified equation (3) above with respect to θ to find the maximum value for θ, and then set it equal to zero to solve for θ. dA/dθ = 36 d/dθ (sinθcosθ) = 36(cosθcosθ + sinθ(-sinθ)) = 36(cos²θ-sin²θ) dA/dθ = 0 => 36(cos²θ-sin²θ) = 0 cos²θ-sin²θ = 0 cos²θ =sin²θ cosθ = sinθ θ = 45º or 135º
In the unit circle, at 45º and 135º angles are where cosθ = sinθ (x = y) that would fit the requirement of the triangle (sum of the 3 angles must equal to 180 degree). But θ is only half of the angle made by AB and AC and 2*(135º) = 270º > 180º, therefore the angle of 135º is impossible and thus leaving us with 45º for θ. angle BAC = 2θ = 2(45º) = 90º
This make the isosceles

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