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March 18th, 2017, 11:46 AM  #1 
Newbie Joined: Jan 2017 From: Toronto Posts: 29 Thanks: 0  Calculating optimal isosceles trapezoid
Trough is to be formed by bending up two sides of a long metal rectangle so that the crosssection 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 
Math Team Joined: Jul 2011 From: Texas Posts: 2,298 Thanks: 1129 
$A = \dfrac{h}{2}(b_1+b_2)$ $h = x\sin{\theta}$ $b_1 = 2(1x)$ $b_2 = 2(1x) + 2x\cos{\theta}$ 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 18th, 2017, 01:07 PM  #3 
Senior Member Joined: Jun 2015 From: England Posts: 509 Thanks: 130 
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  
Newbie Joined: Jan 2017 From: Toronto Posts: 29 Thanks: 0 
Simple and beautiful. I need more practices. Quote:
 
March 19th, 2017, 09:38 AM  #5 
Newbie Joined: Jan 2017 From: Toronto Posts: 29 Thanks: 0 
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... 
March 19th, 2017, 10:57 AM  #6 
Math Team Joined: Jul 2011 From: Texas Posts: 2,298 Thanks: 1129 
$A = \dfrac{x\sin{t}}{2}\bigg[4(1x)+2x\cos{t}\bigg]$ $A = \sin{t}\bigg[2x2x^2+x^2\cos{t}\bigg]$ $\dfrac{dA}{dx} = \sin{t}\bigg[2  4x + 2x\cos{t}\bigg]$ note $\sin{t} \ne 0$ ... $\sin{t}\bigg[2  4x + 2x\cos{t}\bigg] = 0 \implies 2  4x + 2x\cos{t} = 0 \implies x = \dfrac{1}{2\cos{t}}$ $A = \dfrac{x}{2}\bigg[4(1x)\sin{t} + x \cdot 2\sin{t}\cos{t}\bigg]$ $A = \dfrac{x}{2}\bigg[4(1x)\sin{t} + x\sin(2t)\bigg]$ $\dfrac{dA}{dt} = \dfrac{x}{2}\bigg[4(1x)\cos{t} + 2x\cos(2t)\bigg]$ note $x \ne 0$ ... $4(1x)\cos{t} + 2x\cos(2t) = 0$ $4\cos{t}  4x\cos{t} + 2x(2\cos^2{t}1) = 0$ $2\cos{t}  2x\cos{t} + x(2\cos^2{t}1) = 0$ 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  
Newbie Joined: Jan 2017 From: Toronto Posts: 29 Thanks: 0 
You sir are a miracle worker! I finally got the answer. Quote:
 

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calculating, isosceles, optimal, trapezoid 
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