Monday Math Madness #20: Winner!
05 Dec 2008 Quan Quach 7 comments 35 views
MMM #20 Winner
The winner for this edition of MMM is Diego Vila Cid. It appears that Diego has some MATLAB experience as his answer came in the form of a MATLAB script. In fact, his script was eerily reminiscent of how I checked the answer. Kudos, Diego.
This was actually a pretty interesting problem as there were many approaches that were taken in solving it. The most elegant approach was probably done by Richard Berlin. He stated that:
Ignoring the taper on the sides of the pie (which we have no information to calculate) suggests that Rob should take one of the side pieces.
The area on top of the pie is 50*50*PI = 7854. The 26-meter center section is bounded by a rectangle which is 26*100 = 2600 square meters; it’s actually somewhat less than that because of the curvature of the pie.
7854/3 = 2618, so the center section must contain less than a third of the pie.
Nice job Richard. That is an approach that any engineer would be proud to have taken.
Other Solutions
There were mainly two other methods used. One of the solutions employed the use of trig to figure out the area of each section through a clever use of triangles and sectors. The other method used calculus by first finding an equation to represent the circle and then integrating on the limits.
Stay tuned for next week’s problem. I’m sure that Sol will prepare an interesting problem over at wildaboutmath.com this coming Monday.
7 Responses to “Monday Math Madness #20: Winner!”
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Include MATLAB code in your comment by doing the following:
<pre lang="MATLAB">
%insert code here
</pre>
Darn it! Didn’t get the prize. ^_^ But it was a nice problem.
Oh, and for a bit of cruel irony, I found out there was the simpler way, as shown above, the second right after I sent my solution. ^_^
Would it be possible to see the calculus solution?
Here’s the calculus from Kate Nowak:
I wrote a function for the upper semicircle of your diagram y = sqrt(50^2-x^2) and integrated it from x = 13 to x = 50 (using the numerical integration function on a graphing calculator - I could do the trig substitution, but, yuck.). Doubling that, I got the area of a side piece to be 2641.8 m^2. Doubling that and subtracting it from the area of the circle, the area of the middle piece is 2570.4 m^2. So he would want to choose a side piece. I assumed the shape of the solid pie was a cylinder - equal height from the middle to the edges. This would be a nice little problem for a calc 1 course - thanks!
D’oh!
I took the trigonometry path and was estimating the area of the bounding rectangle in mind several times to prove myself but didn’t set this in connection to the third of the whole area. So near, but to complicated thinking.
[...] has announced the winner for MMM #20. Congratulations, Diego Vila [...]
Oooh, I feel famous now. And embarrassed for taking the calculator shortcut.