## Optimization Problem: Rain Gutter

Limits, differentiation, related rates, integration, trig integrals, etc.
bbsa3412
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### Optimization Problem: Rain Gutter

Hey there Purplemath! I need some help setting up the problem:

A rain gutter is to be constructed from a long 30 cm wide sheet of metal by bending two sides upwards at and angle of theta so that each of the sides and bottom is 10 cm wide. Determine the angle theta so that the cross-sectional area of the channel is maximum.

I drew a trapezoid with the long side (l1) at the top and the outside angle formed by the bottom of the trapezoid as theta. I have also labeled the bottom of the trapezoid as l2 and the height as h. I need help setting an equation to relate the area of the gutter with the angle theta. Not quire sure how to proceed. Thanks in advance!

nona.m.nona
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### Re: Optimization Problem: Rain Gutter

A rain gutter is to be constructed from a long 30 cm wide sheet of metal by bending two sides upwards at and angle of theta so that each of the sides and bottom is 10 cm wide. Determine the angle theta so that the cross-sectional area of the channel is maximum.
If I understand correctly, the set-up is along these lines:

Code: Select all

gutter's cross-section: \ | | / \ | | / \|______|/@____
The "@" represents the angle "theta" by which you are to fold up the sheets from the horizonal. The vertical lines and the shorter "base" form right angles, so you can name the angle "inside" as:

$\frac{pi}{2}\, -\, \theta\, =\, \alpha$

You are then trying to find the angle $\alpha$ which maximizes the area of the trapezoid. (You would then back-solve to find the value of $\theta.$)

The height $h$ of the trapezoid is given by:

$\cos(\alpha)\, =\, \frac{h}{10}$

The length of the longer "base" is given by:

$10\, +\, 2b$

where:

$\sin(\alpha)\, =\, \frac{b}{10}$

See where this leads. If you get stuck, please reply showing how far you have gotten. Thank you.