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Old 09-08-2003, 05:27 PM   #1 (permalink)
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constructing functions

Anyone know this math subject pretty well? If so I have a few questions.
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Old 09-08-2003, 05:50 PM   #2 (permalink)
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This is extremely vague. Perhaps post some details?
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Old 09-08-2003, 06:22 PM   #3 (permalink)
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The one in particular I'm havng problems with goes like this.

A wire 10 meters long is to be cut into 2 pieces. One piece will be shaped as an equilateral triangle, and the other piece will be shaped as a circle.

( the top quarter of the wire is 4x the bottom 3/4 is 10-4x)

a) Express the total area ,A, enclosed by the pieces of wire as a function of the length,x, of a side of the equilateral triangle.

I just need to know how to set it up. Everything else I can do. Thanks.
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Old 09-08-2003, 07:47 PM   #4 (permalink)
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this part: ( the top quarter of the wire is 4x the bottom 3/4 is 10-4x) doesn't make any sense to me. as for hte rest of it, well, it's a related rates problem, isn't it?

so, what's the area of a circle? A = πr^2
what about the circumfrence? C = 2*π*r
what's the area of an equilateral triangle? A = 1/2*x*h, where h = the height of the triangle
sin60(degrees) = h/x, so h = xsin60,
so A = 1/2*x*xsin60.
now the trick is to relate all of those, so that you eliminate all the variables besides x.

first step: A = π*r^2 + 1/2*x*x*sin60. only radius isn't meaningful right now, so get rid of using the circumference equation. that's how I'd approach it, anyway. remember that the circumference of your circle will be 10 - 3x, where x is the length of a side of your triangle!
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Old 09-08-2003, 08:09 PM   #5 (permalink)
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The 4x and 10-4x part makes no sense because as I was typing my eyes decided to jump to another question. I do that from time to time. Thanks for the help.
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Old 09-09-2003, 04:53 PM   #6 (permalink)
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Cheers, the base of the triangle is not x, it is 1/3 x (assuming that the other length is 10-x).

If I worked it out right, the final equation is:

(3+(2)(pi)(sin60))(x^2)/12(pi) - 5x/pi + 25/pi

using the quadradic equation, one could then work out the max/min of the equation to maximize the area.
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