Question 159242
A rectangular field is to be enclosed and divided into two sections by a fence
parallel to one of the sides using a total of 600 m of fencing. What is the
maximum area that can be enclosed and what dimensions will give this area? 
:
Let x = the length of the rectangle
Let w = the width
:
Two sections are required, therefore 3w of fence required
:
2x + 3w = 600
also we can write it:
3w = {600-2x)
w = {{{((600-2x))/3}}}
w = 200 - {{{2/3}}}x
:
Find the area;
A = x*w
Replace w with above expression:
A = x(200 - {{{2/3}}}x)
A = 200x - {{{(2/3)x^2}}}
This is a quadratic equation; if we find the axis of symmetry we will know
what value of x gives maximum area:
;
Axis of symmetry formula: x = -b/(2a)
In our equation, a=(-2/3); b=200; therefore;
:
x = {{{(-200)/(2*(-2/3))}}}
x = {{{(-200)/(-4/3))}}}
x = {{{-200 * (-3/4)}}}
x = +150 meters is the length that give max area
:
Find the max area by substituting 150 for x in the area equation
A = 200(150) - {{{2/3}}}*150^2
A = 30000 - 15000
A = 15,000 sq/m is the max area
:
The dimensions of the rectangle: 150 by 100 meters [2(150) + 3(100) = 600]