SOLUTION: John wants to fence a 150 square meters rectangular field. He wants the length and width to be natural numbers {1,2,3,...}. What field dimensions will require the least amount of f

Question 182159: John wants to fence a 150 square meters rectangular field. He wants the length and width to be natural numbers {1,2,3,...}. What field dimensions will require the least amount of fencing?
Found 2 solutions by ankor@dixie-net.com, solver91311:
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
John wants to fence a 150 square meters rectangular field. He wants the length and width to be natural numbers {1,2,3,...}. What field dimensions will require the least amount of fencing?
;
Max area for a given perimeter is a square, therefore
s =
s = 12.247 ft
Using natural numbers: square field 12 by 12 (only 48 m of fencing)
But that is only 144 sq/m
:
If you have to have 150 sq/m and the dimensions have to be integers,
15 by 10 would be my answer. This requires 50 m of fencing
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

I'm not sure how rigorous a treatment of this problem you need. It can be proven that for a given area of a rectangle, the perimeter is minimum when the rectangle is a square.

So, for a 150 square foot area, the minimum perimeter would be a square with side length:

The problem is, is irrational and therefore not a natural number.

Your problem is then to consider all of the 2-factor natural number factorizations of 150.

The prime factorization of 150 is , so the possible 2-factor factorizations are:

Now you could calculate the perimeters for all of these configurations, but remembering that the limiting shape is a square, you just have to find the set of dimensions that are closest to each other in value, namely:

John

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