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Find the smallest perimeter and the dimensions for a rectangle with an area of 225 in^2

User Gruszczy
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To solve this, let the dimensions be x and y. We know that the area is 225. So, x * y = 225


Let the perimeter be denoted as P.


P = 2x + 2y


sine y = 225/x


P = 2x + 450/x


P' = 2 - 450/x^2


P'' = + 900/x^3


Put P' = 0


then x = 15


at x =15 P'' is positive


So the minimum perimeter, will be:
P = 2 (15 + 15) = 60 cm

User Mark Otaris
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