Question 1095575: The area of a rectangular field is 1000 yd2.Two parallel sides are fenced with aluminum at $15/yd. One of the remaining sides is fenced with steel at $10/yd, and all but 10 yd of the remaining side is fenced with wood costing $5/yd.The remaining 10 yd are left unfenced.The total cost of the fencing is $1525. a. What is the length of the side fenced with steel? b. What is the length of each side fenced with aluminum?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! The area of a rectangular field is 1000 yd2.
Two parallel sides are fenced with aluminum at $15/yd.
One of the remaining sides is fenced with steel at $10/yd, and
all but 10 yd of the remaining side is fenced with wood costing $5/yd.
The remaining 10 yd are left unfenced.
The total cost of the fencing is $1525.
:
a. What is the length of the side fenced with steel?
let L = the length fenced with aluminum
let w = the fence with steel
then
L* w = 1000 sq/yds
and
15(2L) + 10w + 5(w-10) = 1525
30L + 10w + 5w - 50 = 1525
30L + 15w = 1525 + 50
30L + 15w = 1575
simplify, divide eq by 15
2L + w = 105
w = 105 - 2L
Using the area equation, replace w with (105-2L)
L(105-2L) = 1000
A quadratic equation
-2L^2 + 105L - 1000 = 0
we can use the quadratic formula; a=-2;b=105, c=-1000; but this will factor
(2L - 25)(L - 40) = 0
two solutions
2L = 25
L = 12.5 yds
and
L = 40 yds, this is the reasonable solution for length using aluminum
:
then
w = 105 - 2(40)
w = 25 yds is fenced with steel
:
:
Check this by finding the total cost with these values for L and w
15(2*40) + 10(25) + 5(25-10) =
1200 + 250 + 75 = 1525
|
|
|
