Homework Help: Optimization Problem

1. Nov 12, 2005

IgniteFlare

Hello,

I am having a little trouble understanding what the question is asking so I was hoping someone would be able to clear up the language the textbook uses. Thanks!

A piece of window framing material is 6m long. A carpenter wants to build a frame for a rural gothic style window where triange ABC is equilateral. The window must fit inside a space 1m wide and 3m high.

a. Find the dimensions that should be used for the 6 pieces so that the maximum amount of light will be admitted. Assume no waste of material for corner cuts, etc.

What I've done so far:
I am assuming that the window width is smaller than the length of BC.
The window framing material totals to 36m.

Thanks for the help!

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2. Nov 12, 2005

sruthisupriya

try this

width=1m,ab=bc=ac=1m.bc is the base.then the height of the traingle is
.5^2+x^2=1(PT).then the height of the rest of the window is3-the answer.tyhen just follow thw usual way.

3. Nov 13, 2005

HallsofIvy

Assuming the picture you show is a picture of the window, then the window width is equal to BC! Since "The window must fit inside a space 1m wide and 3m high", BC must be less than or equal to 1 m.

Also, you are told the "piece of window framing material is 6m long". I don't see where you get 36m.

Call the base x, the height y, and length of the sides of the triangular top z. Then the total length of framing used is 2x+ 2y+ 2z= 6 so x+ y+ z= 3.

The "maximum amount of light will be admitted" when the area of the figure is maximum. The area of the rectangular bottom is xy, of course.

Dropping a perependicular from the top of the triangle divides it into to right triangles with hypotenuse z and base x/2. By the Pythagorean theorem the height of the triangle is $\sqrt{z^2-\frac{x^2}{4}}$ and so the area of the triangle is
$$\frac{x}{2}\sqrt{z^2- \frac{x^2}{4}}$$

The total area of the window is
$$xy+\frac{x}{2}\sqrt{z^2- \frac{x^2}{4}}$$
You want to maximize that subject to the constraint that x+y+z= 3.

Since you haven't shown how you attempted to maximze this, I have no idea what level this is at or what you know about optimizing. There are several different ways to continue but I don't know which to suggest.

4. Nov 13, 2005

IgniteFlare

sruthisupriya, thanks for helping me clear that up a bit.

Sorry about that. How I find the maximum/minimum of an object is to use the derivative and the slope equals to zero.

i.e. (This is how I've been finding max)
y=x^2 + 24x
dy/dx=2x + 24
0=2x + 24

...and you know the rest.

I am trying to attempt the problem this way, so let me know what I am doing wrong:

perimeter of window=6m
6=2L+2W
3=L+W
W=3-L

finding L

A=LW
A=3L-L^2
dA/dL=3-2L
0=3-2L
-3=-2L
1.5=L

Last edited: Nov 13, 2005