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