SOLUTION: A rancher wants to use 400 feet of fencing to enclose a rectangular area of 5100 square feet. What dimensions should the rectangle have?

Question 935989: A rancher wants to use 400 feet of fencing to enclose a rectangular area of 5100 square feet. What dimensions should the rectangle have?
Found 3 solutions by josgarithmetic, lwsshak3, MathTherapy:
Answer by josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!
Dimensions x and y and are unknown.
A and p are known area and perimeter.

-
xy=A

, you can probably factor the quadratic member when the values for p and A are used, but you do not always know. Examples vary.

I leave the rest of the work and decisions to you.
Answer by lwsshak3(11628)   (Show Source): You can put this solution on YOUR website!
A rancher wants to use 400 feet of fencing to enclose a rectangular area of 5100 square feet. What dimensions should the rectangle have?
***
let x=width of rectangular area
length=(400-2x)/2=(200-x)
length*width=area
x(200-x)=5100
200x-x^2=5100
x^2-200x+5100=0
(x-170)(x-30)=0
x=170
or
x=30
200-x=170
What dimensions should the rectangle have?
width=30 ft
length=170 ft

Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!

A rancher wants to use 400 feet of fencing to enclose a rectangular area of 5100 square feet. What dimensions should the rectangle have?

Let length be L, and width, W
Since perimeter = 2(L + W), then: 2(L + W) = 400
2(L + W) = 2(200)
L + W = 200_____W = 200 � L -------- eq (i)
Since area = 5,100 sq ft, and area = LW, then it can be said that: L(200 � L) = 5,100

(L - 170)(L - 30) = 0
L � 170 = 0 OR L - 30 = 0
L, or length = 170 feet OR L = 30
If length = 170, then width = 200 � 170, or 30 feet
If length = 30, then width = 200 � 30, or 170 feet
Therefore, dimensions of rectangle are:
You can do the check!!
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