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Optimization Word Problem

A playing field is being constructed as a rectangle with semi-circles at each end. A 400m racetrack is to be built around the playing field.

a) What radius of the semicircles end would give the playing field it's maximum area: I got r= 200/pi
b) If the track must provide for a straight greater than or equal to section for the 100 m sprints but be less than or equal to 60 m wide, what radius will provide the greatest area for the playing field?

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Brianna L. | Math Education Student -- I love math, and so can you!Math Education Student -- I love math, a...
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Hi Lena,
First off, great job on Part A. I'm also getting r=200/pi. Unfortunately, we must note that this actually gives a rectangle length of 0 -- uh oh! This does make mathematical sense, since a circle is the shape that encloses the greatest area given a certain perimeter, but it doesn't make practical sense for an actual field, and so we go to Part B.
Please note that it's easier for me to just use the symbol for pi when I write, but in the font that this page uses, it apparently just looks like an n: π. Watch out for that :P
We're going to bring some things with us from Part A:
  • the radius of the semicircle is represented by r.
  • the width, or the sides of the field with a semicircle attached, is represented as w, which can be rewritten as w=2r.
  • the length, or the "bare" sides of the field, is represented as l. It can be written in terms of r based on the perimeter: 400=2l + 2πr → l = 200-πr.
  • the area function, which can be expressed as A = (area of rectangle) + (area of semicircles) = (l*w) + (area of one whole circle) = (l*w) + (πr2); then bring in the rewrites of w and l that we just did, and put the whole thing in terms of r to get : ((200-πr)*2r) + (πr2), and simplify to get A= 400r-πr2. Whew!
Alright. Now the wording for Part B is a little funny -- but I'm interpreting it as, the straight length of the track (the rectangle length we found as 0 in the previous part) must be at least 100m, and the rectangle width, also known as twice the circle radius, must be less than or equal to 60m. That would make for a relatively normal-shaped track. Am I reading it right?
This gets a little more complex than Part A, as we have several constraints to work with. Let's look at them:
  • the perimeter being 400m, like in Part A (400 = 2l +πw, rewrite as 400 = 2l + π(2r), rewrite as l = 200-πr).
  • the requirement that the length of the track must be 100m or more (l ≥ 100).
  • the requirement that the width of the track must be 60m or less (w ≤ 60, rewrite as 2r ≤ 60, rewrite as r ≤ 30).
  • the real-life requirements that both r and l must be >0; this is covered by the l ≥ 100 constraint, but we do need r > 0.
And so when we have a multi-constraint problem, we can graph the constraints altogether -- you just have to make sure that your axes are the two quantities that are flexible (in this case the rectangle sides since that's what's changing, but again I'm changing all my w's to 2r's to make it easier on us). Note that you can graph inequalities as shaded regions with a single border line. The perimeter constraint gets graphed as a regular line, since it's just "=" and not "≤" or "≥", and it contains both r and l. See all this in the (quick hand-drawn) graph here:
You know that the maximization, whatever it is, will be a point (r,l) somewhere ON the line l=200-πr, otherwise our 400m perimeter won't happen, right? So we find the piece of this line that falls within our other constraints. It's between r=0 (noninclusive) and r=30 (inclusive) **see footnote for clarification if needed**.
Great -- now we have a range of one variable. So we can look at our final function, the one we want to optimize, the area function -- also in terms of r! -- in that range. Remember from your first problem that area is defined as the function A = 400r-πr2. (Quick note: since we're now finding A from r, instead of l from r, we graph on different axes than before: r is still horizontal, but the vertical axis is now A.) From r=(0,30], the area function looks like this: (this is graphed with x=(0,30) and y=(0,10000)) and we can clearly see that its maximum is all the way to the right, at the point r=30, A=9172.57. 
Check: does this make sense? Let's build that field, with r=30. We get w=60, l=105.75 -- our w and l fit the constraints, and it looks like it's proportioned more like a field. I can also go into the mathematical theory why this is clearly the best solution, if you'd like -- just ask :)
Whew! That took a while to write out, but I think probably you already understood most of it (rewriting the area formulas and such) so hopefully it shouldn't be too tedious to understand. Good luck and as always don't hesitate to ask any followup questions!
Brianna L.
* Why only between r=(0,30]? Because any piece of the line that's outside that range fails some other constraint, like r ≤ 30 or l ≥ 100. This is the only "zone" that we can use.