The cross-section of a drain is a trapezoid. The sides and the bottom of the trapezoid each have a length of 10 feet. Determine the angle θ such that the drain will have maximal cross-sectional area.
For lack of specificity, I will let θ be the angle between the upper base and a side. I expect it will end up 90°, making a square cross-section, but let's wee what happens.
The height of the trapezoid will be 10sinθ, so the area is
a = 10sinθ (10+10+2*10cosθ)
= 10sinθ (20+20cosθ)
= 200sinθ (1+cosθ)
da/dθ = 200cosθ (1+cosθ) + 200sinθ(-sinθ)
= 200(cos^2θ + cosθ - sin^2θ)
= 200(2cos^2θ+cosθ-1)
da/dθ = 0 when cosθ = 1/2 or -1.
Looks like the max area occurs at θ=π/3
Looks like my gut feeling was wrong. Good thing, as it would have been a boring trapezoid.
Oops. I forgot to divide by 2 when getting the average of the bases, but that does not affect the answer in this case.
To determine the angle θ that will maximize the cross-sectional area of the drain, we can use calculus and find the critical points. Since the given information implies that the trapezoid is isosceles, we can consider the drain as a triangle with two equal sides of 10 feet.
Let's use the diagram below to explain the calculation:
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h | \
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10ft 10ft
Here, h represents the height of the triangle, which is the perpendicular distance from the base to the top vertex, and θ is the angle between one of the 10ft sides and the base.
To determine the angle θ that maximizes the cross-sectional area, we need to find the value of θ at which the derivative of the area function is equal to zero or does not exist. This will give us the critical points where the slope of the area function changes.
The formula for the area of a triangle is A = (1/2) * base * height. In this case, the base is the bottom side of the triangle, which is 10ft, and the height is h.
So, the area function can be written as A = (1/2) * 10ft * h = 5h.
Now, to find the value of h in terms of θ, we can use trigonometry. In a right triangle, the opposite side is h, the adjacent side is 5ft (half the base), and the hypotenuse is the side of length 10ft.
Using the sine function, we have sin(θ) = opposite / hypotenuse.
So, sin(θ) = h / 10ft.
Rearranging the equation, h = 10ft * sin(θ).
Now, substitute this expression for h into the area function: A = 5h = 5 * 10ft * sin(θ) = 50ft * sin(θ).
Now, we can take the derivative of the area function with respect to θ to find its critical points:
dA/dθ = 50ft * cos(θ).
Set the derivative equal to zero and solve for θ:
50ft * cos(θ) = 0.
cos(θ) = 0.
θ = π/2.
Since the derivative of the area function is a constant 50ft, which is always positive, there is no non-existence of the derivative. Hence, θ = π/2 radians or 90 degrees is the only critical point.
To confirm that this critical point gives the maximum cross-sectional area, we need to evaluate the second derivative of the area function. If the second derivative is negative at the critical point, it would mean that the critical point is a maximum.
Taking the second derivative of the area function:
d²A/dθ² = -50ft * sin(θ).
Substituting θ = π/2 radians into the second derivative:
d²A/dθ² = -50ft * sin(π/2) = -50ft.
Since the second derivative is negative, this confirms that θ = π/2 radians or 90 degrees is the angle that maximizes the cross-sectional area of the drain.
Hence, the desired angle θ such that the drain will have maximal cross-sectional area is 90 degrees or π/2 radians.