Quadratic Equations 2

[waloshin]

It seems my recent post has been moved to the wasteland already: https://forums.macrumors.com/threads/679777/

So in a follow up am i doing this right?

2) A farmer wishes to fence off a rectangular plot of land along the side of a river. Since the river is very deep, he does not need to build a fence along the river. If he has 600 m of fencing, what dimensions should he use in order to maximize the area of the field What is the maximum area.

x + y = 600
y = 600 - x

a= x(600-x)
a= 600x-x^2

a= -1
b= 600
c= 0

-600
2(-1)


4(-1)(0)-600^2
4(-1)

-360000
-4

90000

vertex ( 300, 90000)

 

[swiftaw]

Since the river is one side of the rectangle, the fence has to cover the other 3 sides.

So x+2y = 600

 

[darklyt]

You're trying to maximize area? A regular polygon maximizes the amount of area bounded by a given perimeter for any number of sides (and a circle maximizes the bounded area for a given perimeter). Since you're looking for a rectangle, the solution is going to be a square. You only have to give three sides of this square with the fourth side the river, so you need three equal lengths of fence. Therefore, we have 600/3 = 200. Thus largest area is 40,000 square feet.

 

[swiftaw]

You're trying to maximize area? A regular polygon maximizes the amount of area bounded by a given perimeter for any number of sides (and a circle maximizes the bounded area for a given perimeter). Since you're looking for a rectangle, the solution is going to be a square. You only have to give three sides of this square with the fourth side the river, so you need three equal lengths of fence. Therefore, we have 600/3 = 200. Thus largest area is 40,000 square feet.

Not true.

Let the rectangle be x by y. Then, since the fence has to cover 3 sides, we have that x+y+y = 600, or x = 600-2y.

Thus the area of the rectangle = x*y = (600-2y)*y = 600y - 2y^2

Thus dA/dy = 600 - 4y

Solving for zero gives us y = 150, thus x = 600-2*150 = 300

Thus the max area is 300*150 = 45000

 

[darklyt]

Not true.

Let the rectangle be x by y. Then, since the fence has to cover 3 sides, we have that x+y+y = 600, or x = 600-2y.

Thus the area of the rectangle = x*y = (600-2y)*y = 600y - 2y^2

Thus dA/dy = 600 - 4y

Solving for zero gives us y = 150, thus x = 600-2*150 = 300

Thus the max area is 300*150 = 45000

Nice catch. I should have spent more than a few seconds looking at it lol.

If it were four sides involved, then it would have been necessary to have a square since if P=total perimeter, P/2 * P/2 = (P^2)/4, while (P/2 - x)(P/2 + x) = (P^2)/4 - x^2 < (P^2)/4 for x>0. It is interesting that the solution you found is half a square with sides of length 300. Also interesting is that if it were at a plot of land that lied next to a river that (magically) formed a right angle next to this land, then we would have the solution is a square.