# What value of θ will maximize the trough’s volume?

The trough in the figure is to be made to the dimensions shown. Only the angle θ can be varied.

• 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 + b) * 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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