Question 182159
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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:



*[tex \LARGE \text{          }\math sqrt{150} = 5 sqrt{6}]


The problem is, *[tex \Large 5 sqrt(6)] 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 *[tex \Large 2 \times 3 \times 5 \times 5], so the possible 2-factor factorizations are:


*[tex \LARGE \text{          }\math 2 \times 75]


*[tex \LARGE \text{          }\math 3 \times 50]


*[tex \LARGE \text{          }\math 5 \times 30]


*[tex \LARGE \text{          }\math 6 \times 25]


*[tex \LARGE \text{          }\math 10 \times 15]


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: *[tex \Large 10 \times 15] 


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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