- #1
inevitable08
- 7
- 0
Homework Statement
An open gutter with cross section in the form of a trapezoid with equal base angles is to be made by bending up equal strips along both sides of a long piece of metal 12 inches wide. Find the base angles and the length of the sides for maximum carrying capacity.
For more clarification the part about bending equal strips of a 12 inch wide sheet means... There are 3 sides to the trapezoid (open top) the 2 sides are equal so then we are left with a varying base.
Homework Equations
side length = s
base length = b
[itex]\nabla[/itex]f = [itex]\lambda[/itex][itex]\nabla[/itex]g(s,b,[itex]\theta[/itex])
Area of trapezoid = (1/2)(b1 + b2)h
The Attempt at a Solution
Ok first I believe this would be a function of three variables; s, b and [itex]\theta[/itex]. This is because each one of those will have a factor on whether it fits the constraint and what the area of the trap to be.
The constraint will be:
g(s,b,[itex]\theta[/itex]) = 2*s + b = 12
Now here's the fun part I believe:
Area = (1/2)(b1 + b2)h
if we take a part of the trapezoid that's the triangle and find h in terms of s and [itex]\theta[/itex]:
h = s*cos([itex]\theta[/itex])
now we don't care about b2 since this is a top open gutter right and the constraint calls for only 3 sides of the gutter. So here I figured I would still need b2 i just need it in terms of something else.
b2 = b1 + 2*s*sin([itex]\theta[/itex])
so we substitute b2 and h into the original Area equation:
Area = (1/2)(b1+b1+2*s*sin([itex]\theta[/itex])(s*cos)
now add b1 + b1 = 2b1 and distribute the (1/2) and the s*cos([itex]\theta[/itex]) we end up with:
Area = s*b1*cos([itex]\theta[/itex]) + s2sin([itex]\theta[/itex])cos([itex]\theta[/itex])
whew. So now we can use our double angle formula sin(2[itex]\theta[/itex]) = 2sin([itex]\theta[/itex])cos([itex]\theta[/itex])
Area = s*b1*cos([itex]\theta[/itex]) + (1/2)s2sin(2[itex]\theta[/itex])
*this is so we can take partials easier and don't have to deal with the product rule*
Partials of F(s,b,[itex]\theta[/itex]) = Area, then find critical points:
fs = b1*cos([itex]\theta[/itex]) + s*sin(2[itex]\theta[/itex]) = 0
if s = 0 then b1 = 0 or [itex]\theta[/itex] = pi/2
fb1 = s*cos([itex]\theta[/itex]) = 0
s = 0 or [itex]\theta[/itex] = pi/2
f[itex]\theta[/itex] = -s*b1*sin([itex]\theta[/itex]) + s2cos(2[itex]\theta[/itex]) = 0
s = 0
then I started setting those equal to the g(s,b1,[itex]\theta[/itex]) we have 4 equations 4 unknowns and start solving for various variables and substituting. It got REALLY ugly and then I stopped. Somethings not right... Chances are you haven't even read this far cause you already found a mistake :) which I hope. Cause this is just a ridiculous problem.