##### main algebraic solution

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May 30th, 2015

side of the square = 36/4 =9 inch

Area of square = side x side = 9x9 = 81  sq inch

Let the width of rectangle is  W therefore Length will be 2W

perimeter = 2 ( Length +width ) = 2( 2W+W) = 36 or W = 6 inch

Hence Width = 6 inch and Length = 12 inch

Therefore area of Rectangle = Length x Width = 12x6 =72 sq inch

Hence Area of Square will be lesser than that of Rectangle of same length...

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May 30th, 2015

wrong

May 30th, 2015

Let the square have sides, s and the rectangle have sides r and 2r.
Perimeters:   36 = 4s + 2(r + 2r) = 4s + 6r

s = (1/4)(36 - 6r) = 9 - 1.5r
areas: A = s² + 2r² = (9 - 1.5r)² + 2r² = 81 - 27r + 4.25r²
Then dA/dr = 0 = 8.5r - 27
r = 27/8.5 = 3.17 inch  (Width of rectangle)

Is this a local max, or min? d²A/dr² = 8.5 > 0, so it is a minimum.

Please check now and give me good comment .............

May 30th, 2015

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May 30th, 2015
Width of rectangle    = 3.17 inch       Is correct !!!!!!!!!!!!

May 30th, 2015

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May 30th, 2015
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May 30th, 2015
Oct 23rd, 2017
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