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March 18th, 2017, 11:46 AM   #1
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Calculating optimal isosceles trapezoid

Trough is to be formed by bending up two sides of a long metal rectangle so that the
cross-section of the trough is an isosceles trapezoid. If the width of the
metal sheet is 2 meters, how should it be bent to maximize the volume of the trough?

Answer: The sides and bottom should all be 2/3 meter, and the sides should be bent up at angle pi/3.

Any tips would be much appreciated.
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March 18th, 2017, 12:58 PM   #2
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$A = \dfrac{h}{2}(b_1+b_2)$

$h = x\sin{\theta}$

$b_1 = 2(1-x)$

$b_2 = 2(1-x) + 2x\cos{\theta}$

I started by setting $\dfrac{\partial A}{\partial x} = 0$, solved for $x$ in terms of $\theta$ ...
$x=\dfrac{1}{2-\cos{\theta}}$.

Set $\dfrac{\partial A}{\partial \theta} = 0$, substituted for $x$, found $\cos{\theta} = \dfrac{1}{2} \implies x = \dfrac{2}{3} \text{ and } \theta = \dfrac{\pi}{3}$

Give it a go ...
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File Type: jpg max_trap.jpg (7.5 KB, 1 views)
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March 18th, 2017, 01:07 PM   #3
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These figures will not quite work in real life, however since sheet metalwork requires a bend allowance.
This will affect the base dimension slightly.

SheetMetal.Me – Bend Allowance
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March 18th, 2017, 01:23 PM   #4
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Simple and beautiful. I need more practices.


Quote:
Originally Posted by skeeter View Post
$A = \dfrac{h}{2}(b_1+b_2)$

$h = x\sin{\theta}$

$b_1 = 2(1-x)$

$b_2 = 2(1-x) + 2x\cos{\theta}$

I started by setting $\dfrac{\partial A}{\partial x} = 0$, solved for $x$ in terms of $\theta$ ...
$x=\dfrac{1}{2-\cos{\theta}}$.

Set $\dfrac{\partial A}{\partial \theta} = 0$, substituted for $x$, found $\cos{\theta} = \dfrac{1}{2} \implies x = \dfrac{2}{3} \text{ and } \theta = \dfrac{\pi}{3}$

Give it a go ...
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March 19th, 2017, 09:38 AM   #5
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Here is my work..

P = 2 - 2x
H = x * cos(θ)
Q = 2x * sin(θ) + P = 2 * xsin(θ) + 2 - 2x

A = 1/2 (P + Q) * H
A = 1/2 * (4 - 4x + 2x * sin(θ)) * x * cos(θ)
A = (2 - 2x + x * sin(θ)) * x * cos(θ)

∂A/∂x = 2 * cos(θ) - 4x * cos(θ) + 2x sin(θ) cos(θ) = 0
∂A/∂θ = -2x * sin(θ) + 2x^2 * sin(θ) + 2x * (cos(θ)^2 - sin(θ)^2) = 0

∂A/∂x = cos(θ) - 2x * cos(θ) + x sin(θ) cos(θ) = 0
∂A/∂θ = -x * sin(θ) + x^2 * sin(θ) + x * cos(θ)^2 - x * sin(θ)^2 = 0


2 unknowns, but with 2 complicated equations...
Any tips to help me moving forward would be much appreciated...
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File Type: jpg trapezoid.jpg (4.2 KB, 2 views)
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March 19th, 2017, 10:57 AM   #6
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$A = \dfrac{x\sin{t}}{2}\bigg[4(1-x)+2x\cos{t}\bigg]$

$A = \sin{t}\bigg[2x-2x^2+x^2\cos{t}\bigg]$

$\dfrac{dA}{dx} = \sin{t}\bigg[2 - 4x + 2x\cos{t}\bigg]$

note $\sin{t} \ne 0$ ...

$\sin{t}\bigg[2 - 4x + 2x\cos{t}\bigg] = 0 \implies 2 - 4x + 2x\cos{t} = 0 \implies x = \dfrac{1}{2-\cos{t}}$


$A = \dfrac{x}{2}\bigg[4(1-x)\sin{t} + x \cdot 2\sin{t}\cos{t}\bigg]$

$A = \dfrac{x}{2}\bigg[4(1-x)\sin{t} + x\sin(2t)\bigg]$

$\dfrac{dA}{dt} = \dfrac{x}{2}\bigg[4(1-x)\cos{t} + 2x\cos(2t)\bigg]$

note $x \ne 0$ ...

$4(1-x)\cos{t} + 2x\cos(2t) = 0$

$4\cos{t} - 4x\cos{t} + 2x(2\cos^2{t}-1) = 0$

$2\cos{t} - 2x\cos{t} + x(2\cos^2{t}-1) = 0$

substitute $\dfrac{1}{2-\cos{t}}$ for $x$ ...

$\dfrac{(2\cos{t})(2-\cos{t})}{2-\cos{t}} - \dfrac{2\cos{t}}{2-\cos{t}} + \dfrac{2\cos^2{t}-1}{2-\cos{t}} = 0$

$4\cos{t}-2\cos^2{t} - 2\cos{t} + 2\cos^2{t}-1 = 0$

$2\cos{t} - 1 = 0 \implies \cos{t} = \dfrac{1}{2} \implies t = \dfrac{\pi}{3}$

$x = \dfrac{1}{2-\cos\left(\frac{\pi}{3}\right)} = \dfrac{2}{3}$
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March 19th, 2017, 01:10 PM   #7
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You sir are a miracle worker! I finally got the answer.

Quote:
Originally Posted by skeeter View Post
$A = \dfrac{x\sin{t}}{2}\bigg[4(1-x)+2x\cos{t}\bigg]$

$A = \sin{t}\bigg[2x-2x^2+x^2\cos{t}\bigg]$

$\dfrac{dA}{dx} = \sin{t}\bigg[2 - 4x + 2x\cos{t}\bigg]$

note $\sin{t} \ne 0$ ...

$\sin{t}\bigg[2 - 4x + 2x\cos{t}\bigg] = 0 \implies 2 - 4x + 2x\cos{t} = 0 \implies x = \dfrac{1}{2-\cos{t}}$


$A = \dfrac{x}{2}\bigg[4(1-x)\sin{t} + x \cdot 2\sin{t}\cos{t}\bigg]$

$A = \dfrac{x}{2}\bigg[4(1-x)\sin{t} + x\sin(2t)\bigg]$

$\dfrac{dA}{dt} = \dfrac{x}{2}\bigg[4(1-x)\cos{t} + 2x\cos(2t)\bigg]$

note $x \ne 0$ ...

$4(1-x)\cos{t} + 2x\cos(2t) = 0$

$4\cos{t} - 4x\cos{t} + 2x(2\cos^2{t}-1) = 0$

$2\cos{t} - 2x\cos{t} + x(2\cos^2{t}-1) = 0$

substitute $\dfrac{1}{2-\cos{t}}$ for $x$ ...

$\dfrac{(2\cos{t})(2-\cos{t})}{2-\cos{t}} - \dfrac{2\cos{t}}{2-\cos{t}} + \dfrac{2\cos^2{t}-1}{2-\cos{t}} = 0$

$4\cos{t}-2\cos^2{t} - 2\cos{t} + 2\cos^2{t}-1 = 0$

$2\cos{t} - 1 = 0 \implies \cos{t} = \dfrac{1}{2} \implies t = \dfrac{\pi}{3}$

$x = \dfrac{1}{2-\cos\left(\frac{\pi}{3}\right)} = \dfrac{2}{3}$
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