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Find the dimensions of a rectangle with area 2197 m^2 whose perimeter is as small as possible?

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Dr. Juanita Wolff

Feb 20, 2025

Let x be the side length of one side, and y the other side. Then xy = 2197 as given, since area = length x width.Now the perimeter is P = 2(x + y) From the above equation, y = 2197/xthus P = 2(x + 2197/x)we derive to find dP/dx = 2(1 - 2197/x^2) by the power ruleslet dP/dx = 0 ---0 = 2(1 - 2197/x^2)0 = 1 - 2197/x^21 = 2197/x^2x^2 = 2197 x = 13sqrt(13) - we reject the negative solution since dimensions cannot be negative but we are told xy = 2197, so y = 2197/(13sqrt(13)) = 13sqrt(13)Thus the dimension are 13sqrt(13) and 13sqrt(13) - it is a square.... Show More
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