Question

A farmer wishes to enclose a pasture that is bordered on one side by a river (so one of the four sides wont require fencing) She has decided to create a rectangular shape for the area and will use barbed wire to create the enclosure there are 600 feet of wire availible for this project and she will use all the wire. what is the maximum area enclosed by the fence then find the maximum of the function. ...?

Answer 1

The area enclosed is (2x)(600-2x).(2x)(600−2x)=1200x−4x2
In order to maximize we need the derivative to be equal to 0.y′=1200−8x=0
1200=8x
x=150
Therefore the sides for maximum area are 150*300.
The area is: 45000

Answer 2

Answer:

Area is 45000 square feet.

Step-by-step explanation:

Let x be the length of fenced side parallel to the river side.

Let y be the length of other two sides.

Then,

$x+ 2y 600$

So, we get $x=600-2y$

Area of this rectangle = xy

= $y(600-2y)$

$x=-2y^{2}+600y$

Now attached is the graph of the area function, which is a parabola opening downward.

We can see ta the maximum area occurs when y=$-600/[2(-2)]$ = 150

And $x = 600-2(150)$

x =$600-300=300$

Therefore, to maximize the area, the side parallel to the river is to be 300 feet long, and the other two fenced sides should be 150 feet long.

The area is = $300\times150= 45000$ square feet.