MATH 1320 : Spring 2014
Lab Instructor : Kurt VanNess
Bonus Lab Solutions
A long piece of galvanized sheet metal with width w is to be bent into a symmetric form with three straight
sides to make a rain gutter.
1. Determine the dimensions that allow the maximum possible flow; that is, find the dimensions that give
the maximum possible cross-sectional area.
Solution
I will leave it to you to draw a picture of this gutter. Label the bottom length as w − 2x, the two side
lengths as x and the top two angles of the trapezoid as θ. Clearly, θ ≤ 90◦ . This is a maximization
problem, so we need to find a function that represents the area. Using the geometry of the problem,
we can define the area of the symmetric trapezoid by summing the area of the end triangles and the
area of the center rectangle. We have
A(x, θ) = 2(1/2)(x sin θ)(x cos θ) + (x sin θ)(w − 2x)
= x2 sin θ cos θ + xw sin θ − 2x2 sin θ
To find a maximum, we want to find when Ax = 0 and Aθ = 0. So we take the partial derivatives and
set them equal to zero to obtain
Ax (x, θ) = 2x sin θ cos θ + w sin θ − 4x sin θ
= sin θ(2x cos θ + w − 4x) = 0
Aθ (x, θ) = x2 cos2 θ − x2 sin2 θ + xw cos θ − 2x2 cos θ
= x(x cos2 θ − x sin2 θ + w cos θ − 2x cos θ) = 0
Since we know that the maximum doesn’t occur when sin θ = 0 (θ = 0 or θ = π) and when x = 0
(both give areas of 0), then we consider the system of equations
0 = 2x cos θ + w − 4x
0 = x cos2 θ − x sin2 θ + w cos θ − 2x cos θ
Seeing the w alone in the first equation, I think to use substitution over some sort of elimination.
Solving for w we have
w = 4x − 2x cos θ
so then the substituting this into the second equation and simplifying we have
0 = x cos2 θ − x sin2 θ + (4x − 2x cos θ) cos θ − 2x cos θ
= x cos2 θ − x sin2 θ + 4x cos θ − 2x cos2 θ − 2x cos θ
= −x sin2 θ − x cos2 θ + 2x cos θ
= −x(sin2 θ + cos2 θ) + 2x cos θ
= −x + 2x cos θ
= x(2 cos θ − 1)
Since we already said x = 0 gives an area of 0, then we only consider 2 cos θ − 1 = 0. Thus, we know
cos θ = 1/2 or θ = π/3 . We know w = 4x − 2x cos θ, so solving for x we have
x=
w
w
=
4 − 2 cos θ
3
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MATH 1320 : Spring 2014
Lab Instructor : Kurt VanNess
2. Would it be better to bend the metal into a gutter with a semicircular cross-section?
Solution
Since we are trying to maximize the cross-sectional area, we need to find the radius of the semicircle
we can make with the sheet metal of width w. The sheet metal will cover half the circumference of a
full circle so we have
(1/2)2πr = πr = w
Therefore, r = w/π. This would yield a gutter with cross-sectional area
A=
1 2
1 w 2
w2
πr = π
=
≈ 0.1592w2
2
2
π
2π
The cross sectional area we found with the previous conditions is
A = x2 sin θ cos θ + xw sin θ − 2x2 sin θ
!
√ !
w 2
w 2 √3 1 w 3
w
−2
=
+
3
2
2
3
2
3
√ 2
√ 2
√ 2
3w
3w
3w
+
−
=
36
6
9
√ 2
3w
≈ 0.1443w2
=
12
√ !
3
2
So it would be better to bend the metal into a gutter with a semicircular cross-section.
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