Create a rectangle with a perimeter that is half as large as the area

Question

Create a rectangle with a perimeter that is half as large as the area 

Answer 1

If we call the side lengths of the rectangle X and Y, then the area will be:

 

A = XY

 

and the perimeter will be:

 

P = 2X + 2Y

 

If we want the perimeter to be 1/2 as large as the area, then

 

P = 1/2*A

 

2X + 2Y = 1/2*XY

 

4X + 4Y = XY

 

We can write this as:

 

XY - 4X = 4Y

 

X*(Y - 4) = 4Y

 

X = 4Y/(Y - 4)

 

So this gives you a general formula for a rectangle's side lengths that are required to make this rectangle fit the description you want. It doesn't seem to work for any side length Y that is less than or equal to 4. Plug in any value of Y you want that's greater than 4 and it'll spit out a value for X that will make your rectangle work.

 

Say we chose Y = 5.

 

X = 4*5/(5 - 4)

X = 20

 

So the area is X*Y = 100

 

And the perimeter is 2(X+Y) = 2(20+5) = 50

 

So you can see the perimeter is "half as large" as the area, even though the units of area and perimeter are different.

 

If we chose Y = 10;

 

X = 40/(10-4) = 40/6 = 20/3

 

So the area will be:

 

XY = 200/3

 

Perimeter will be:

 

2*(10 + 20/3) = 2*(30/3 + 20/3) = 100/3

 

Seems to check out, even though its use is limited.

 

This is the best I could come up with given the constraints. Hope this helps.