2400 yards of fencing to enclose a rectangular area. find the dimensions that maximize the enclosed area

Question

pre-calculus questions that looks for I believe the minimum and maximums to identify the length and width

Answer 1

Set the length to be x and width to be y.

 

So 2x+2y = 2400, y = 1200 - x

 

Area = xy = x(1200-x) = - x² +1200x = -(x-600)²+360000

 

From this expression. we can conclude that the Maximum area of 360000 yard² can only be achieved when x = 600. (other values of x will make the -(x-600)² a negative term that reduces the total area, the area will be 0 if x = 0, and that will be the minimum are that you can achieve here)

 

Therefore, x = 600, y = 600 are the dimensions that would maximize the enclosed area. 

 

 

Answer 2

P=2l+2w

2400=2l+2w

1200=l+w

1200-l=w

A=lw

A=l(1200-l)

A=1200l-l²

use -b/2a

-1200/-2=600 yards on a side will maximize the area; therefore you have a square 600 by 600

simply speaking, the maximum area will always be a square

(before showing your work !!!) you can get the answer by dividing 2400/4=600