A rectangle with an area of 48 square units, what is the largest perimeter you can have?
It seems that the bigger the difference between the length and the width, the bigger the perimeter. So if the length and width had to be integers, the largest perimeter you could get would be 98 (48 by 1 rectangle). But you could keep halving one side of the rectangle, and doubling the other, and the perimeter would just keep getting bigger and the area would remain the same. Maybe it's worth noting that the perimeter is getting bigger by a value converging to 2, but I have no idea if that has any significance.
Having a length and a width that are closer to each others values just makes the perimeter smaller and makes it more "squarish". I'm running into problems when I try and explain why a bigger difference in length and width makes the perimeter bigger with solid maths.
if the area = ab
and perimeter = 2a+2b
then ab=48 and b=48/a
2a+2(48/a)
2a+(96/a)= perimeter
The smaller the value of a the bigger the value of 96/a, and the bigger the value of a the bigger the value of 2a.
This isn't sufficient as you can plug in any value and the same rules still apply. I'm having difficulty explaining myself.
The smallest perimeter would be obviously be if the shape was a square, making the length of each side √48, leaving the perimeter as 16√3 ≈ 27.71
As one of the sides lengths "move away" from that value (the other sides length obviously changing with respect to the other, keeping the area as 48 square units) the perimeter gets bigger and bigger.
Lets just say that a = 12
2a = 24
96/a = 8
Okay, lets increase a by 4, so a = 16
2a = 32
96/a = 6
Note how 2a increases by a much bigger amount, than the amount 96/a decreases by. In fact, 2a increases by an amount 4 times bigger than the amount 96/a decreased by. If it 2a increases faster than 96/a decreases, then surely the perimeter would be getting bigger.
I'm still working on the problem, I will update when I have anything new to say.
