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Please help! Find the dimensions of a rectangle with area 1000 m2 whose perimeter is as small as possible.?

2 Answers

E
Everardo Feil

Feb 20, 2025

Area = L W ( L = length ; W = width)1000 = L WL = 1000 / WPerimeter = 2L+2WP = 2(1000/W) + 2WP = 2000 / W + 2WdP/dW = -2000 /W^2 + 2 = 0-2000 /W^2 = -22W^2 = 2000W^2 = 1000W = sqrt(1000) = 31.63L = 1000 / W = 1000 /31.63 = 31.63The dimensions are L = 31.63 m and W = 31.63 md^2P/dW^2 = 4000 / w^3 > 0 , so the perimeter is minimum... Show More
E
Everardo Feil

Feb 20, 2025

It’s a square of side √1000 = 31.623 m (A square has the largest area to perimeter ratio of all rectangles)Proof: P = 2a + 2b, where b = 1000/a, so P = 2a + 2000/a = 2a + 2000a^(-1) (Notice how to write exponents)Differentiate and set derivative to zerodP/da = 2 – 2000a^(-2) = 2 – 2000/a^2 = 0, giving 2a^2 = 2000 from which a = √1000 = b... Show More

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