SOLUTION: For the given area, find the length and width of the rectangle with the least perimeter. Use whole numbers only. 30 m2 40 yd2

Algebra ->  Algebra  -> Surface-area -> SOLUTION: For the given area, find the length and width of the rectangle with the least perimeter. Use whole numbers only. 30 m2 40 yd2      Log On

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Question 194543This question is from textbook Harcourt Math
: For the given area, find the length and width of the rectangle with the least perimeter. Use whole numbers only.
30 m2
40 yd2This question is from textbook Harcourt Math

Answer by Alan3354(8189) About Me  (Show Source):
You can put this solution on YOUR website!
For the given area, find the length and width of the rectangle with the least perimeter. Use whole numbers only.
30 m2
40 yd2
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A square has the least perimeter for a rectangle. (A proof is available is needed.)
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For squares, P = 4*s
30 sq meters --> sides of sqrt(30) meters
l = w = sqrt(30) = ~ 5.477 meters
Using integers, --> 5 by 6.
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40 sq yds --> sqrt(40)
l = w = sqrt(40) = ~ 6.325 yards
Using integers --> 5 by 8 yards
This is a silly problem.