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?

Coolio

lw=1000
lets assume that the aluminum sides are the legnth
2 sides
2*15=30
30l=cost of aluminum

steel=10w
wood=(w-10)5=5w-50
we can eliminate the w by solving for w in first relation

lw=1000
divide both sides by l
w=1000/l
sub that for w

10(1000/l)=steel
5(1000/l)-50=wood

wood cost+steel+aluminum cost=total cost
10(1000/l)+5(1000/l)-50+30l=1525
(10000/l)+(5000/l)-50+30l=1525
(15000/l)-50+30l=1525
add 50 to both sides
(15000/l)+30l=1575
times both sides by l
15000+30l²=1575l
minus 1575l from both sides
30l²-1575l+15000=0
factor
(15)(l-40)(2l-25)=0
set each equal to 0
l-40=0
l=40

2l-25=0
2l=25
l=12.5

the there are 2 possible combinations

aluminum=40yd and steel=25yd or
aluminum=12.5yd and steel=80ft
both yield 1000yd² and cost $1525