Find the dimensions of a rectangle with area 27000 m2 whose perimeter is as small as possible......?

Question

Find the dimensions of a rectangle with area 27000 m2 whose perimeter is as small as possible. (Give your answers in increasing order, to the nearest meter.)

Answer 1

Let L (meters) be the length of this rectangle. The width of this rectangle is: W = (27000/L) meter. Also let l be the square root of L, that is: l = sqrt(L). The parameter of this rectangle is thus:
P = 2(L + 27000/L)
= 2{ l^2 - 2*sqrt(27000) + 27000/l^2 + 2*sqrt(27000) }
= 2*(l - 30*sqrt(30)/l)^2 + 4*sqrt(27000)
The first term 2*(l - 30*sqrt(30)/l)^2 is a square, and takes its minimum value Zero and thus P gets minimized ONLY when L = l^2 = 30*sqrt(30)m. Hence: When L = 30*sqrt(30)m, and W = 30*sqrt(30)m, ie, when the rectangle is actually a square, P gets minimized to be 120*sqrt(30) m.

Answer 2

enable "L" be the size. in view that length circumstances width provides you the section, this suggests the width is 2700 / L. the fringe is the sum of all 4 ingredient lengths: L + L + 2700/L + 2700/L = 2L + (5400/L) we could desire to locate the fee of "L" which will shrink this. So take the by-product, set the expression equivalent to 0, and resolve for L. you may desire to finally finally end up with 30?3.