Question

A plastic gutter is designed to catch water at the edge of a roof.

Manufacturers need to minimize the amount of material used to make their product. What is the best cross-section for a gutter?

Options for cross-section are:
1. Triangular Shape
2. Rectangular Shape
3. Circular Shape

I need to use optimization and calculus to solve this. Any working out you can show me would be very helpful.

Answer 1

Here is the exact task for anyone who wants more info:

Answer 2

I haven't done this stuff in quite some time but I think I can help you still.

First what equations do you need to utilize?

Answer 3

no equations are given, only what is written in the attachment.

Answer 4

Yes but you can't do optimization without equations

Answer 5

These are pretty common equations

Answer 6

basically, i need to propose a value for the volume of each cross section, then optimize from there.

Answer 7

"Manufacturers need to minimize the amount of material used to make their product."

what are the manufactures interested in optimizing?

Answer 8

these are elementary geometry equations

Answer 9

ok, all I know is that which is on the task sheet in the attachment. I have been given no further information. After reading the task sheet, you and I will be on the same level of understanding.

Answer 10

You need to use equations for surface area and volume because you are interested in optimizing the surface area to volume ratio

Answer 11

the manufacturer wants to use the least amount of material to achieve the most volume

Answer 12

which means i need to maximize the volume of the (lets start with number 1) triangle, but minimum TSA of the two sides. So say I propose a volume of 100 cubed units....

Answer 13

You need to relate both equations then take the first and second derivatives to determine the functions minimums and maximums depending on what relation you set your equation up

Answer 14

to have

Answer 15

I have to go now. I wont be back on for a while but I'll continue work on the question. Thanks for the help so far, I'll get back to you on this later.

Answer 16

ok whatever