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 March 18th, 2017, 11:46 AM #1 Senior Member   Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3 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. 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 ...
Attached Images max_trap.jpg (7.5 KB, 2 views) March 18th, 2017, 01:07 PM #3 Senior Member   Joined: Jun 2015 From: England Posts: 915 Thanks: 271 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 March 18th, 2017, 01:23 PM   #4
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Simple and beautiful. I need more practices.

Quote:
 Originally Posted by skeeter $A = \dfrac{h}{2}(b_1+b_2)$ $h = x\sin{\theta}$ $b_1 = 2(1-x)$ $b_2 = 2(1-x) + 2x\cos{\theta}$ 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 ... 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...
Attached Images trapezoid.jpg (4.2 KB, 2 views) March 19th, 2017, 10:57 AM #6 Math Team   Joined: Jul 2011 From: Texas Posts: 3,031 Thanks: 1620 $A = \dfrac{x\sin{t}}{2}\bigg[4(1-x)+2x\cos{t}\bigg]$ $A = \sin{t}\bigg[2x-2x^2+x^2\cos{t}\bigg]$ $\dfrac{dA}{dx} = \sin{t}\bigg[2 - 4x + 2x\cos{t}\bigg]$ 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 $x$ ... $\dfrac{(2\cos{t})(2-\cos{t})}{2-\cos{t}} - \dfrac{2\cos{t}}{2-\cos{t}} + \dfrac{2\cos^2{t}-1}{2-\cos{t}} = 0$ $4\cos{t}-2\cos^2{t} - 2\cos{t} + 2\cos^2{t}-1 = 0$ $2\cos{t} - 1 = 0 \implies \cos{t} = \dfrac{1}{2} \implies t = \dfrac{\pi}{3}$ $x = \dfrac{1}{2-\cos\left(\frac{\pi}{3}\right)} = \dfrac{2}{3}$ 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 $A = \dfrac{x\sin{t}}{2}\bigg[4(1-x)+2x\cos{t}\bigg]$ $A = \sin{t}\bigg[2x-2x^2+x^2\cos{t}\bigg]$ $\dfrac{dA}{dx} = \sin{t}\bigg[2 - 4x + 2x\cos{t}\bigg]$ 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 $x$ ... $\dfrac{(2\cos{t})(2-\cos{t})}{2-\cos{t}} - \dfrac{2\cos{t}}{2-\cos{t}} + \dfrac{2\cos^2{t}-1}{2-\cos{t}} = 0$ $4\cos{t}-2\cos^2{t} - 2\cos{t} + 2\cos^2{t}-1 = 0$ $2\cos{t} - 1 = 0 \implies \cos{t} = \dfrac{1}{2} \implies t = \dfrac{\pi}{3}$ $x = \dfrac{1}{2-\cos\left(\frac{\pi}{3}\right)} = \dfrac{2}{3}$ December 4th, 2017, 09:52 PM #8 Newbie   Joined: Dec 2017 From: China Posts: 2 Thanks: 0 You can find other explanations in this post: https://machinemfg.com/press-brake-ultimate-guide/ Last edited by skipjack; December 5th, 2017 at 03:16 AM. Tags calculating, isosceles, optimal, trapezoid Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post mobel Probability and Statistics 1 May 24th, 2016 09:22 AM get42n8 Algebra 6 January 5th, 2015 11:21 AM kittycub4 Calculus 2 October 21st, 2011 03:55 PM Recipe Algebra 4 March 11th, 2010 07:32 PM dai_lo Calculus 1 February 27th, 2008 11:45 AM

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