Question 333063
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I did a little analysis using a spread sheet.


Your parameters are:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2l\ +\ 2w\ =\ lw]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ l\ \in\ \mathbb{Z},\ l\ >\ 0]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w\ \in\ \mathbb{Z},\ w\ >\ 0]


From the first equation we can derive:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ l\ =\ \frac{-2w}{2\ -\ w}]


From which we can restrict *[tex \Large w > 2]


Consider the function


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(l,w)\ =\ 2l\ +\ 2w\ -\ lw]


Let *[tex \Large l\ =\ 1] and let *[tex \Large w] take on the values 3, 4, 5, and 6.  Then *[tex \Large f(1,w)] takes on values of 5, 6, 7, and 8 and will increase without bound as *[tex \Large w] increases without bound.  Clearly *[tex \Large l\ =\ 1] is not a possibility since we are looking for a result of *[tex \Large f(1,w)\ =\ 0]


Let *[tex \Large l\ =\ 2] and let *[tex \Large w] take on the values 3, 4, 5, and 6.  Then *[tex \Large f(2,w)] has a constant value of 4 and will remain 4 as *[tex \Large w] increases without bound.  *[tex \Large l\ =\ 2] is not a possibility.


Let *[tex \Large l\ =\ 3] and let *[tex \Large w] take on the values 3, 4, 5, and 6.  Then *[tex \Large f(3,w)] takes on values of ... of ... wait for it... 3, 2, 1, and ZERO!  We found one.  *[tex \Large f(3,w)] continues to decrease without bound as *[tex \Large w] increases without bound.  Hence there will be no other values of *[tex \Large w] such that *[tex \Large f(3,w)\ =\ 0]


Let *[tex \Large l\ =\ 4] and let *[tex \Large w] take on the values 3, 4, 5, and 6.  Then *[tex \Large f(4,w)] takes on values of 2, 0, -2, -4, decreasing without bound as *[tex \Large w] increases without bound.  Another possibility found and another guarantee that it is the only possibility for *[tex \Large l\ =\ 4].


Let *[tex \Large l\ =\ 5] and let *[tex \Large w] take on the values 3, 4, 5, and 6. Then *[tex \Large f(5,w)] takes on values of 1, -2, -5, -8, decreasing without bound as *[tex \Large w] increases without bound -- and skipping right over zero.  No luck here.


Let *[tex \Large l\ =\ 6] and let *[tex \Large w] take on the values 3, 4, 5, and 6. Then *[tex \Large f(6,w)] takes on values of 0, -4, -8, -12, decreasing without bound as *[tex \Large w] increases without bound.  Another possibility found and another guarantee that it is the only possibility for *[tex \Large l\ =\ 6].


Notice that for *[tex \Large l\ =\ 6], we found a possibility when *[tex \Large w] was at its low limit of 3.  What do you suppose is going to happen when we let *[tex \Large l\ =\ 7]?


Right.  Let *[tex \Large l\ =\ 7] and let *[tex \Large w] take on the values 3, 4, 5, and 6. Then *[tex \Large f(7,w)] takes on values of -1, -6, -11, -16, decreasing without bound as *[tex \Large w] increases without bound.  No zero here, or will there be one as far as you care to go.


Now, Let *[tex \Large w\ =\ 3] and let *[tex \Large l] take on the values 8, 9, 10, and 11.  Then Then *[tex \Large f(l,3)] takes on values of -2, -3, -4, -5, decreasing without bound as *[tex \Large l] increases without bound.  You should be convinced by now that we won't find any more instances of *[tex \Large f(l,w)\ =\ 0]


Drop me a note and I'll share a graph of this.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
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