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?
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.
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).
.
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.
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.