Question

Find the dimensions of a rectangle with area 729000 m2 whose perimeter is as small as possible. (Give your answers in increasing order, to the nearest meter.)

Answer 1

Explanation:

Area of a rectangle = Length * Width

A = LW

729000 = LW .... 1

Perimeter of a rectangle P = 2L+2W

P = 2L+2W

From 1;

W = 729000/L

P = 2L + 2(729000/L)

P = 2L + 1458000/L

To minimize the perimeter, dP/dL = 0

dP/dL = 2 - 1458000/L²

2 - 1458000/L² = 0

2L² - 1458000 = 0

2L² = 1458000

L² = 729000

L = √729000

Length = 853.8m

W = A/L

Width = 729000/853.8

Width = 853.8

Hence the dimension to the nearest meter is 854m by 854m