The Parkhursts used 160 yd of fencing to enclose a rectangular corral to divide it into two parts by a fence parallel to one of the shorter sides. Find the dimensions of the corral if its area is 1000 yd ^2 ( yd has a small 2 above the d)
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I answered this already
The the width be x yds
let the length be y yds
then 3x + 2y = 160
y = (160-3x)/2
area = xy
x(160-3x)/2 = 1000
160x - 3x^2 = 2000
3x^2 - 160x + 2000 = 0
(x-20)(3x - 100)= 0
x = 20 or x= 100/3
if x=20, then y = 50
if x=100/3 then y = 30
Case 1, x=20, y =50
3(20) + 2(50) = 160, area = 20*50 = 1000
Case 2 x = 100/3, y =30
3(100/3) + 2(30) = 160, area = 30*100/3 = 1000
So the dimensions could be 20 by 50, with the third side parallel to the 20
or
the dimensions could be 100/3 by 30 with the third side parallel to the 100/3 side.
To find the dimensions of the corral, we need to set up an equation based on the given information.
Let's assume that the length of the corral is "L" and the width is "W." We know that the total amount of fencing used is 160 yards, and we can express this as an equation:
2L + W = 160 Equation 1
We are also given that the area of the corral is 1000 square yards, and we know that the area of a rectangle is calculated by multiplying its length by its width:
L * W = 1000 Equation 2
Now, we have a system of two equations with two unknowns (L and W). To solve this system, we can use substitution or elimination.
Let's solve it using substitution. Rearrange Equation 2 to solve for L:
L = 1000 / W
Substitute this value of L into Equation 1:
2(1000 / W) + W = 160
Now, we can simplify this equation:
2000 / W + W = 160
Multiply through by W to eliminate the fraction:
2000 + W^2 = 160W
Rearrange this equation and set it equal to zero:
W^2 - 160W + 2000 = 0
Now, we can either use factoring, completing the square, or the quadratic formula to solve for W.
In this case, let's use factoring to find the values of W. The quadratic equation can be factored as:
(W - 40)(W - 50) = 0
This means that either (W - 40) = 0 or (W - 50) = 0. Therefore, we have two possible values for W:
W1 = 40 or W2 = 50
Now, substitute these values of W back into Equation 1 to find the corresponding values of L:
For W1 = 40:
2L + 40 = 160
2L = 120
L = 60
For W2 = 50:
2L + 50 = 160
2L = 110
L = 55
So, we have two possible sets of dimensions for the corral:
1) Length (L) = 60 yards, Width (W) = 40 yards
2) Length (L) = 55 yards, Width (W) = 50 yards
Therefore, the dimensions of the corral can be either 60 yards by 40 yards or 55 yards by 50 yards.