A rectengular area is to be enclosed on three sides by fencing, an existing hedge forming the fourth side. Find the greatest area of the rectangle and its dimensions if 140mof fencing is available.

Answer 1

let the length parallel to the hedge be y
let the width be x

so y + 2x = 140
y = 140-2x

area = xy
= x(140-2x) = -2x^2 + 140x

if you know Calculus:
-4x + 140 = 0
x = 35
then y = 140 - 70 = 70

max area = 35(70) = 2450

If no Calculus:
complete the square to find the vertex

A = -2x^2 + 140x
= -2(x^2 - 70x)
= -2(x^2 - 70x + 1225 - 1225)
= -2( (x-35)^2 - 1225)
= -2(x-35)^2 + 2450
so the vertex is (35,2450)
giving us a max of 2450 when x = 35

Answer 2

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