Math
posted by Nkosingiphile .
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.

let the length parallel to the hedge be y
let the width be x
so y + 2x = 140
y = 1402x
area = xy
= x(1402x) = 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( (x35)^2  1225)
= 2(x35)^2 + 2450
so the vertex is (35,2450)
giving us a max of 2450 when x = 35 
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