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# Carl has a rectangle whose side lengths are positive integers. This rectangle has the property that when he increases the width by 1 unit an

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Carl has a rectangle whose side lengths are positive integers. This rectangle has the property that when he increases the width by 1 unit and decreases the length by 1 unit, the area increases by x square units. What is the smallest possible positive value of x?

Guest Oct 26, 2014

#3
+85624
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Note that...for any given perimeter, the area is at a max when L = W.

So let L=W and so that LW = max area

Now, if we decrease W by 1 and increase L by 1, we have

(W - 1) (L +1) = WL + (W-L) -1  = max area + 0 - 1  = max area - 1

And if  we decrease W by 2 and increase L by 2, we have

(W - 2)(L+2) = WL + 2(W-L) - 4  = max area + 0 - 4 = max area - 4

And with successive increases/decreases of n = 3, 4, 5.......etc., the max area is decreased by n^2. So, the smallest of these decreases occurs between n = 0 and n = 1. Or, looking at it from another perspective, inceasing the Width from (W -1) to W and decreasing the Length from (L+1) to L ( where W = L), results in x = 1.

CPhill  Oct 26, 2014
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#1
+26625
+5

Original rectangle  L*W = A

New rectangle   (L-1)*(W+1) = A + x

L*W + L - W - 1 = A + x

or L - W - 1 = x   since L*W = A

If x has to be a positive integer as well then:

When L-W = 2 units,  x = 1 square unit

If x doesn't have to be an integer then there it has no smallest value (it can be infinitesimally small).

.

Alan  Oct 26, 2014
#2
+92193
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It increases the area by x^2  Alan.

sorry

I was thinking it increases the area by x^2 units squared.

but

you have interpreted it as x units squared.

Your interpretion is probably the intended one.

Melody  Oct 26, 2014
#3
+85624
+5

Note that...for any given perimeter, the area is at a max when L = W.

So let L=W and so that LW = max area

Now, if we decrease W by 1 and increase L by 1, we have

(W - 1) (L +1) = WL + (W-L) -1  = max area + 0 - 1  = max area - 1

And if  we decrease W by 2 and increase L by 2, we have

(W - 2)(L+2) = WL + 2(W-L) - 4  = max area + 0 - 4 = max area - 4

And with successive increases/decreases of n = 3, 4, 5.......etc., the max area is decreased by n^2. So, the smallest of these decreases occurs between n = 0 and n = 1. Or, looking at it from another perspective, inceasing the Width from (W -1) to W and decreasing the Length from (L+1) to L ( where W = L), results in x = 1.

CPhill  Oct 26, 2014

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