A rectangle if made from 50 feet of fencing .what is the maximum area that can be enclosed?explain & show work

Question

please explain and show all work

Answer 1

The maximum area would be a circle. If it has to be a rectangle a square would be the highest area. So divide 50 by 4 and you get 12.5 and that will be the largest. It is a square 12.5 feet per side.

Answer 2

The rectangle with the greatest area will be the one with equal sides.

So if each side is defined by lengths A,B C & D

and A = B = C = D

and A+B+C+D = 50

then A= 12.5
B=12.5
c=12.5
and d=12.5

Answer 3

The maximum area is 156.25 square feet. Using simple logic, you will have each side of the rectangle be 12.5 feet.

50/4 = 12.5

12.5 * 12.5 = 156.25

Answer 4

150 sq ft
Rectangle is 10 ft x15 ft
2 sides x 10 ft = 20 ft
2 sides x 15 ft=30 ft
therefore you have a total of 50 ft fencing

Answer 5

let W be the width of the rectangle
let L be the length of the rectangle

50 = 2W + 2L
L = 25 - W

and the area is
A = WL
A = W(25-W)
A = 25W - W^2

differentiate and set to zero
dA/dW = 25 - 2W == 0
W = 25/2 = 12.5
L = 25 - 12.5 = 12.5

Hence a square with sides of 12.5 satisfies your requirements.

Answer 6

OK:.. maximizing areas is an application for derivatives.

You know the perimeter of you rectangle
50 ft=2x+2y when x is the length and y is the height

Get everything on terms of x:
2y=50-2x
y=(50-2x)/2
y=25-x

You want to maximize your area
MAXIMIZE=x*y

Substitute y in the equation for maximization:
MAX=x*(25-x)
MAX=25x - x^2

Now you get this equations derivative:
MAX' = 25 - 2x

You get the roots of this equation:
25-2x=0
2x=25
x=25/2

You got your critical point. (x=12.5, y=12.5)
Your area would be 156.25 ft^2

Cynthia

Answer 7

625 sq feet, 25x25