What value of θ will maximize the trough’s volume?
What value of θ will maximize the volume of the trough? The trough, as depicted in the accompanying figure, is designed with specific dimensions. The only variable that can be adjusted is the angle θ.
3 Answers
In order to do this, we only really need to know when the trapezoid has the greatest area
9 * cos(t) is the height
4 + 2 * 9 * sin(t) is the 2nd base
A = (1/2) * (b[1] + b[2]) * h
A = (1/2) * (4 + 4 + 18 * sin(t)) * 9 * cos(t)
A = (9/2) * (8 + 18 * sin(t)) * cos(t)
A = (9/2) * 2 * (4 + 9 * sin(t)) * cos(t)
A = 9 * (4 + 9 * sin(t)) * cos(t)
A = 9 * (4 * cos(t) + 9 * sin(t) * cos(t))
A = 9 * (4 * cos(t) + (9/2) * sin(2t))
dA/dt = 9 * (-4 * sin(t) + 9 * cos(2t))
dA/dt = 0
0 = 9 * (-4 * sin(t) + 9 * cos(2t))
0 = 9 * cos(2t) - 4 * sin(t)
4 * sin(t) = 9 * cos(2t)
4 * sin(t) = 9 * (1 - 2 * sin(t)^2)
4 * sin(t) = 9 - 18 * sin(t)^2
18 * sin(t)^2 + 4 * sin(t) - 9 = 0
sin(t) = (-4 +/- sqrt(16 + 36 * 18)) / 36
sin(t) = (-4 +/- 2 * sqrt(4 + 9 * 18)) / 36
sin(t) = (-2 +/- sqrt(4 + 162)) / 18
sin(t) = (-2 +/- sqrt(166)) / 18
sin(t) > 0
(sqrt(166) - 2) / 18
sin(t) = (sqrt(166) - 2) / 18
t = arcsin((sqrt(166) - 2) / 18)
t = 37.20525383385804154732087023355 degrees , 0.64935417844106618398071005902938 radians
The height of the trough is 9*cos(theta).
The horizontal leg of each small triangle is 9*sin(theta).
The area of the rectangle is 36*cos(theta).
The area of the two small triangles together is
81*sin(theta)*cos(theta) or 40.5*sin(2*theta).
The total cross-sectional area of the trough is
A = 36*cos(theta) + 40.5*sin(2*theta).
dA/d(theta) = -36*sin(theta) + 81*cos(2*theta).
Setting this expression to 0, we get
36*sin(theta) = 81*cos(2*theta) =>
4*sin(theta) = 9*[cos^2(theta) - sin^2(theta)] =>
4*sin(theta) = 9*[1 - 2*sin^2(theta)] =>
18*sin^2(theta) + 4*sin(theta) - 9 = 0.
Using the quadratic formula, you have
sin(theta) = -1/9 +/- (1/36)*sqrt(16+648) = +0.60467.
theta = 37.2 degrees.
To see that this makes SOME sense, note that if theta = 45 degrees, the cross-section area is 81/2 + 18*sqrt(2) = about 66; if theta = 30 degrees, the cross-section area is 81*sqrt(3)/4 + 18*sqrt(3) = about 66.25; but if theta = 36.87 degrees (giving 3-4-5 triangles), the area is
972/25 + 3.2*9 = around 67.68...i.e., a little bigger than what you get for theta = 30 or 45.
find the Θ which yields max area of the end...if the height is x then the the top is 4 + 2 [ 9 sin Θ ] where
x = 9 cos Θ...thus A = [ base 1 + base 2] / 2 [ height ] = [4 + 9 sin Θ ] [ 9 cos Θ ] =
36 cos Θ + 81 sin Θ cos Θ ====> - 36 sin Θ + 81[ cos² Θ - sin² Θ ] = 0...let w = sin Θ
- 36 w + 81 [ 1 - 2 w² ] = 0 ===> w = you finish
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