Asked by Alexander
4. A rectangular enclosure is formed by using 800m of fencing. Find the greatest possible area that can be enclosed in this way and the corresponding dimensions of the rectangle.
Answers
Answered by
Damon
4 x = 800
x = 200 feet on a side square
=========================
prove that square is max area
A = x y
P = 2 x + 2 y perimeter
so x + y = P/2
A = x (P/2 - x) = -x^2 +P/2 x
x^2 - (P/2) x = A find vertex,complete square
x^2 - (P/2)x + P^2/16 = -A + P^2/16
(x-P/4)^2 = -(A-P^2/16)
vertex at x= P/4 (then y = P/4 too :)
A = (P/4)^2
so square with side = Perimeter/4 is max area
x = 200 feet on a side square
=========================
prove that square is max area
A = x y
P = 2 x + 2 y perimeter
so x + y = P/2
A = x (P/2 - x) = -x^2 +P/2 x
x^2 - (P/2) x = A find vertex,complete square
x^2 - (P/2)x + P^2/16 = -A + P^2/16
(x-P/4)^2 = -(A-P^2/16)
vertex at x= P/4 (then y = P/4 too :)
A = (P/4)^2
so square with side = Perimeter/4 is max area