Maximizing Flow in a Rain Gutter: 3-sided vs Semicircular Cross Section

[clairaut]

A long piece of sheet metal w inches wide is to be bent into a SYMMETRIC form with three straight sides to make a rain gutter. A cross section is shown below.

\_____/

The base is w-2x and the angled side lengths are both x with a theta between top horizontal.

A. Determine dimensions that allow maximum possible flow or maximum cross sectional area.

B. Would it be better to bend the material into a gutter with a semicircular cross section than a three sided cross section?

 

[clairaut]

I can't seem to find a constraint. Without constraint, I cannot use lagrangian method.

So, I take the partials of the cross sectional area function with respect to x and then with respect to theta.

I get nowhere afterwards.

 

[HallsofIvy]

First this is clearly a home-work type question and should have been posted there.

Second the "constraint" is x= w/4.

Frankly I wouldn't consider this a "Lagrange multiplier" problem to begin with. I would write the volume of the trapezoid as a function of x and \theta, replace x by w/4, and treat it as a "Calculus" I problem of finding the value of the single variable \theta that maximizes the area.

 

[clairaut]

Why would constraint be w/4

 

[clairaut]

Also, volume of trapezoid is not important here. I only need the cross sectional area of this trapezoid.

 

[clairaut]

?

 

[verty]

Show some working out please. Give us the formula for the cross-sectional area, it'll have 3 variables: w, x, Θ. If you can do that, we can start to help.

 

[clairaut]

[A(x,b)] = (x^2-b^2)^.5 (w + b - 2x)

Where b is the leg of each of the right triangles in the corners. [A(x,@)] = (w)(x)[sin(@)] + (x^2)[sin(@)][cos(@)] - 2x^2[sin(@)]

Above are the cross sectional surface area equations.

 

[verty]

[A(x,b)] = (x^2-b^2)^.5 (w + b - 2x)

Where b is the leg of each of the right triangles in the corners. [A(x,@)] = (w)(x)[sin(@)] + (x^2)[sin(@)][cos(@)] - 2x^2[sin(@)]

Above are the cross sectional surface area equations.

I get the same formula although you should try to simplify it a little. Okay, you want to find the maximum. Can you see how to do it? It will of course have something to do with the gradient. Refer to your textbook if you need to.

 

[clairaut]

Yes. No problems with gradient. It's the CRITICAL POINTS that are hard to find.

 

[verty]

I've helped all I can, sorry.

 

[clairaut]

Lol. How have you helped?

You've merely agreed with me... No more

 

[micromass]

It's the CRITICAL POINTS that are hard to find.

Please post an attempt. Also, this thread belongs in homework, so please post there in the future.