algebra question?

Question

What is the number of units in the positive difference between the maximum and minimum possible perimeters of a rectangle with whole number dimensions and area 72 square units?

please explain! thank you thank you

Answer 1

The smallest possible perimeter will be when the side lengths are nearest to the root of 72 (or closest together)

sq(72) = 8.48....

8 and 9 are nearest. The perimeter is 2(8) + 2(9) = 34

The largest possible perimeter is when the values are farthest apart. In this case, that would be 72 and 1. The perimeter is 2(72) + 2(1) = 146.

146-34 = 112. That is your answer.

Answer 2

Well, it turns out that the perimeter is minimum when the rectangle is as close to a square as possible. That would be when one side is 9 and the other side is 8. Then the perimeter would be 2x9 + 2x8 = 18+16 = 34.

The perimeter gets much larger the more stretched out the rectangle gets. So, the worst you could do would be rectangle where one side is 1 and the other side is 72. then the perimeter would be 2x1 + 2x72 = 2 + 144 = 146.

The difference between these two is 146 - 34 = 112.

Answer 3

lb=72
min perimeter when l=b or l^2=72 => l=sqrt72 <9
so since dimensions are whole numbers the closest are l=9, b=8
or perimeter=2l+2b =2(8+9)=34units

maximum is when one of the sides is reduced to smallest whole number i.e. b=1, so since lb =72 therefore l = 72
hence perimeter = 2*73 =146
the positive difference is : 146-34 =112 units

Answer 4

Area s=x*y; x & y being sides of rectangle; p=2x+2y is its perimeter; y=s/x;
Thus p=2x+2*s/x, so p=p(x) and p’(x)=2-2s/x^2. Now if p’(x)=0 we can find extrema;
p’=0=2*(1-s/x^2) or x1=sqrt(s)=sqrt(72)=8.49 –is not integer, let’s round it to nearest x1=8, then y1=9, as 8*9=72; thus p1=16+18=34;
Now lim(p)=lim[2*(x+s/x)] {for x->0} =infinity, but we have condition for x being integer; thus x=1, then y=72; thus p2=2+144=146; therefore p2-p1=146-34=112.