SOLUTION: 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 t
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-> SOLUTION: 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 t
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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?
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
