which dimensions produce the smallest perimeter for a rectangular area.

Question

a rectangular area of 120m^2 

Answer 1

Proven many times:

The rectangle with the greatest area and the smallest perimeter is the square.

Each side of the square is 2√30.

 

Answer 2

Whoot whoot! An optimization problem! I don't see what math class you're in. I'm going to use Calc 1.
Let me know if that does not work for you.

"smallest" means minimal. So we want to minimize the perimeter in meters
area = LW

p = 2L+2w 

120 = LW

Now let's write with one variable only. Doesn't matter which one, I chose to work with W's.

L = 120/W

p = 2(120/W) +2W

p = 240/W +2W
Tada, one variable!

 

Let's write the expression using exponents:
p = 240W^-1 +2w

Let's find critical points for this equation.

To do so, take the derivative and set it =0.

dp/dw = -240W^-2+2 

0 = -240w^-2 +2

-2 = -240w^-2

-2/-240=w^-2

1/120 = 1/w2

W= square root of 120 =10.95

If you want to prove that as the minimum, use the first derivative test.
Pop in a value less than 10.95 and a value greater than 10.95 into the first derivative dp/dw.
Avoid all the math, just compute whether the value would be positive or negative.
Where the value of dp/dw is negative, the function is decreasing. 
Where the value of dp/dw is positive, the function is increasing.
Do it and you'll see.

So the value of W is 10.95 that minimizes the perimeter.
L = 120/w

L = 120/10.95

L= 10.95

 

Both L and W are 10.95, so we're looking at a square: 10.95m x10.95m

 

 

Answer 3

Hi Jamal!  I am not sure what course you are working on.  This is a typical problem in a Calculus 1 course, but I can show you a simpler way to get to the answer.  

 

Consider some different rectangles that have an area of 120.

 

One such rectangle has dimensions 1 x 120.  This makes the perimeter = 1 + 120 + 1 + 120 = 242

Another rectangle has dimensions 2 x 60. This makes the perimeter = 2 + 60 + 2 + 6 = 124 (smaller)

Another rectangle has dimensions 3 x 40.  The perimeter = 3 + 40 + 3 + 40 = 86 (even smaller)

 

4 x 30 gives a perimeter of 68

5 x 24 gives a perimeter of 58 

6 x 20 gives a perimeter of 52

8 x 15 gives a perimeter of 46

 

Notice the perimeter keeps getting smaller as the two dimensions get closer and closer to being the same size.

We might conclude (correctly) that if the length and the width of the rectangle were the same, this would give us the smallest perimeter.

 

If the length and width of the rectangle (having area 120) were the same, their lengths would be the square root of 120.  

.

 

Hope this helps!  :)

Simplifying the sqrt(120), we get 2 times the sqrt(30).   This makes the perimeter 8 times the sqrt(30)