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Julie's garden with perimeter of 54 ft and width of 12 ft is to have posts put around it 3 feet away from the garden's edge on all sides. They are also to be three feet apart. how many posts will she need?

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closed as off-topic by El-Guest, JMP, Bass, Glorfindel, Quintec Sep 8 '18 at 23:06

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – El-Guest, JMP, Bass, Glorfindel, Quintec
If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ I think it does work, but this doesn’t really remind me of a puzzle so much as it does a word problem from a set of math homework. I might vote to close this because of that; and if closed I will remove my answer. $\endgroup$ – El-Guest Sep 8 '18 at 12:40
  • $\begingroup$ I agree with @El-Guest as this is just a normal math homework that I used to get from school. Voting to close this :D $\endgroup$ – Kevin L Sep 8 '18 at 13:00
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Julie’s garden

Will be $15 \times 12$ yielding a $54$ foot perimeter. If we add 3 feet to all sides, we have a length of $21$ feet and a width of $18$ feet. This means we’ll need 8 posts for the length and 7 posts for the width; and then we need to deduct 4 posts since we counted the corners twice.

The answer is therefore

$8+8+7+7-4 = 26$ posts.

@Jaap Scherphuis has raised an excellent extension in the comments below. We ought to round the corners such that we minimize the perimeter.

The resulting fence shape looks like a running track, albeit more square than that. We must compute the total perimeter. We see that the shape has to have straightaway stretches of 15 feet and 12 feet for the lengths and widths respectively. At each corner, we create a quarter circle of radius 3 away from the corner. The total perimeter of this shape is therefore $54 + 6\pi$ feet. Spacing each post 3 feet apart would therefore require $18 + 2\pi \approx 24$ posts, although each post wouldn’t be exactly 3 feet apart from each other.

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  • $\begingroup$ Is there no way to cheat at the corners? It is not a given that the there must be posts exactly at the corners of the expanded rectangle, which are in fact much more than 3 foot from the perimeter of the garden. $\endgroup$ – Jaap Scherphuis Sep 8 '18 at 13:50
  • $\begingroup$ @JaapScherphuis So then what you’re truly looking for is some sort of rounded oval I suppose — you’re absolutely right about this and I’d be very interested in seeing the OP’s thoughts on this, though I fear it’s outside the scope of their intended question. $\endgroup$ – El-Guest Sep 8 '18 at 13:55
  • $\begingroup$ Even if you do keep all the posts on the expanded rectangle (i.e. don't round the corners), you can still save some fencing by not putting the posts at the corners. I don't think it adds up to a full 3 feet's worth though, so you probably can't do it with one fewer post in that way. $\endgroup$ – Jaap Scherphuis Sep 8 '18 at 14:02
  • $\begingroup$ You’d have to do some shifting around the corners though, because taking out the corner post leaves the two posts nearest the corner $3\sqrt{2}$ feet apart I believe. $\endgroup$ – El-Guest Sep 8 '18 at 14:05
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    $\begingroup$ @Jay look at the second solution in my answer. There’s a trade off between 3 feet of distance between the posts and the garden, and 3 feet of space between the posts and each other. There is no way to have both. $\endgroup$ – El-Guest Sep 8 '18 at 15:21

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