Maximizing area of trapezoid Drain Gutter

[testme]

Homework Statement

Image is very big.

Angles A and B are 120°
length of metal is 5m
width of metal is 0.6m
it's an isosceles trapezoid drainage.

Homework Equations

° in a trapezoid: 180(n-2)
A = 2(bh/2) + wh

The Attempt at a Solution

I'm not sure, I started but I'm not even sure if I'm thinking of it right since the question is stated in a confusing way. This is what I believe I have to do so far.

w = AD + AB + BC = 0.6
l = 2x + 2y = 5
x = z + 2b

A = bh + wh

I know it's really complicated but I'm not quite sure how to go about it, I'm just wondering if someone can tell me exactly what the question is asking and if I'm on the right track, thanks.

 

[Ray Vickson]

Homework Statement

Image is very big.

Angles A and B are 120°
length of metal is 5m
width of metal is 0.6m
it's an isosceles trapezoid drainage.

Homework Equations

° in a trapezoid: 180(n-2)
A = 2(bh/2) + wh

The Attempt at a Solution

I'm not sure, I started but I'm not even sure if I'm thinking of it right since the question is stated in a confusing way. This is what I believe I have to do so far.

w = AD + AB + BC = 0.6
l = 2x + 2y = 5
x = z + 2b

A = bh + wh

I know it's really complicated but I'm not quite sure how to go about it, I'm just wondering if someone can tell me exactly what the question is asking and if I'm on the right track, thanks.

It is impossible to tell if you are on the right track, because you do not define your symbols. For example, in the equation A = bh + wh, is A = area, or is A = angle A, or what? What are b,h,w? In the equation 2x + 2y = 5, what are x and y supposed to represent?

In this problem it is clear that the length 5m is irrelevant; you are asked to maximize the cross-sectional area, subject to bottom + sides adding up to 60 cm.

RGV

 

[testme]

Sorry about that.

A = bh + wh
where b is the b of the triangles if you cut up the trapezoid and h is the height. w is the width of the top after the two triangle bases have been taken out.

So then it would have to be that

top + bottom + 2side = 0.6 m?
the top = bottom + 2base

so then could it be written as

0.6 = 2bottom + 2side + 2base?

 

[Office_Shredder]

Let's assume that AB is the bottom of the trapezoid. Then there is no top! The gutter doesn't have a top to it. So just bottom+2*side = .6

Aso, it's not true that top=bottom+2*base, you need to use a little trigonometry to figure out what the relationship really is

In this problem the only real control that you have is how long AB is - this determines every other aspect of the gutter. So your objective is to write the area function purely as a function of AB by using the constraints in the problem... in this case we have

area=wb+hb

w is just the length of AB, so you need to describe the values of b and h in terms of the length of AB

 

[Ray Vickson]

Sorry about that.

A = bh + wh
where b is the b of the triangles if you cut up the trapezoid and h is the height. w is the width of the top after the two triangle bases have been taken out.

So then it would have to be that

top + bottom + 2side = 0.6 m?
the top = bottom + 2base

so then could it be written as

0.6 = 2bottom + 2side + 2base?

No: look at the figure; the top is open, so no metal is used on the top. You still need to figure out the area.

RGV

 

[testme]

Using w as the length of AB, h as the height made, and b would then be the base. would it be

Area = wh + 2(bh/2)
Area = wh + bh

Then for 0.6 being the width of the entire sheet

0.6 = BC + AB + AD
BC = AD, let x = BC = AD
0.6 = AB + 2x
we'll use w again for AB
0.6 = w + 2x

 

[testme]

I got the right answer, it's not too hard once I understood exactly what they meant and a little bit of a push in the right direction. Though working with sines of angles isn't my favorite. Thanks again :)