Find the dimensions of a rectangle with area 1000 m2 whose perimeter is as small as possible. (Give your answers in increasing order, to the nearest meter.)

### 2 Answers

Area = L W ( L = length ; W = width) 1000 = L W L = 1000 / W Perimeter = 2L+2W P = 2(1000/W) + 2W P = 2000 / W + 2W dP/dW = -2000 /W^2 + 2 = 0 -2000 /W^2 = -2 2W^2 = 2000 W^2 = 1000 W = sqrt(1000) = 31.63 L = 1000 / W = 1000 /31.63 = 31.63 The dimensions are L = 31.63 m and W = 31.63 m d^2P/dW^2 = 4000 / w^3 > 0 , so the perimeter is minimum

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 zero

dP/da = 2 – 2000a^(-2) = 2 – 2000/a^2 = 0, giving 2a^2 = 2000 from which a = √1000 = b