Find the dimensions of a rectangle with area 2197 m^2 whose perimeter is as small as possible?
What are the dimensions of a rectangle with an area of 2197 m² that has the smallest possible perimeter? Please provide a detailed explanation of how to arrive at the answer, including any necessary calculations or formulas used in the process.
2 Answers
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/x
thus P = 2(x + 2197/x)
we derive to find dP/dx = 2(1 - 2197/x^2) by the power rules
let dP/dx = 0 ---
0 = 2(1 - 2197/x^2)
0 = 1 - 2197/x^2
1 = 2197/x^2
x^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.
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