SOLUTION: hey, could you please solve this problem for me. and thank you. A rectangular field is to be enclosed and divided into two sections by a fence parallel to one of the sides us

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Question 159242: hey,
could you please solve this problem for me. and thank you.
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?
i will be waiting for your answer,
and thank you,
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
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 = %28%28600-2x%29%29%2F3
w = 200 - 2%2F3x
:
Find the area;
A = x*w
Replace w with above expression:
A = x(200 - 2%2F3x)
A = 200x - %282%2F3%29x%5E2
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 = %28-200%29%2F%282%2A%28-2%2F3%29%29
x = %28-200%29%2F%28-4%2F3%29%29
x = -200+%2A+%28-3%2F4%29
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%2F3*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]