## Need help with an Area question: "You are to build two separate enclosures..."

Geometric formulae, word problems, theorems and proofs, etc.
brett1977
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Joined: Sun Aug 23, 2015 6:02 am
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### Need help with an Area question: "You are to build two separate enclosures..."

1. You have been given two lengths of fencing.
2. One is exactly 246.54918439354108 meters in length; the other is exactly 393.5680673801993 meters.
3. You are to build two separate enclosures - one enclosure for each of the lengths of fencing.
4. The area of each enclosure must be the maximum obtainable from each length of fencing.
5. The more significant figures, the more accurate your results.
6. Round final results to 3 decimal points.

If anyone could help it would be amazing!

buddy
Posts: 197
Joined: Sun Feb 22, 2009 10:05 pm
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### Re: Need help with an Area question: "You are to build two separate enclosures..."

1. You have been given two lengths of fencing.
2. One is exactly 246.54918439354108 meters in length; the other is exactly 393.5680673801993 meters.
3. You are to build two separate enclosures - one enclosure for each of the lengths of fencing.
4. The area of each enclosure must be the maximum obtainable from each length of fencing.
5. The more significant figures, the more accurate your results.
6. Round final results to 3 decimal points.
Are you allowed to use the side of the bigger one as part of the smaller one or do they have to be completely discrete? I think probably discrete. so what did you get when you did your equations for each one (like they show here)? Like for the 1st one you did 2L + 2W = 246.54918439354108 and LW = A. You solved the 1st eqn & plugged into the 2nd. then you solved. What did you get? Etc. Thnx.

Alfred
Posts: 36
Joined: Fri Mar 14, 2014 6:44 pm
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### Re: Need help with an Area question: "You are to build two separate enclosures..."

1. You have been given two lengths of fencing.
2. One is exactly 246.54918439354108 meters in length; the other is exactly 393.5680673801993 meters.
3. You are to build two separate enclosures - one enclosure for each of the lengths of fencing.
4. The area of each enclosure must be the maximum obtainable from each length of fencing.
5. The more significant figures, the more accurate your results.
6. Round final results to 3 decimal points.
Remarks:
why two enclosures...both work out the same way...
why "the more sig. figures the better" if you're to round out to 3 decimals only?!
Anyhow:
P = 246.54918439354108
Max area of enclosure with perimeter P is a CIRCLE, so (R = radius, A = area):
P = pi(2R) : R = P / (2pi) : A = pi(R^2)

That'll come out to A = 4837.2360000000009629798...amazing!!!
So 4837.236 to 3 decimal points.
SURE looks like a circle was intended!!

Now, you can do the other one yourself, right?
It'll come out to 12326.2180000000000133054....amazing again!!