SOLUTION: can you please help me with this word problem. thank you very much 112. Swimming pool design. An architect has a designed a motel pool with in a rectangular area the is fenced

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Question 59822: can you please help me with this word problem. thank you very much
112. Swimming pool design. An architect has a designed a motel pool with in a rectangular area the is fenced on three sides. If she uses 60 yards of fencing to enclose as area of 352 square yards, then what are the dimensions for L and W? Assume L is greater than W
Found 2 solutions by checkley71, joyofmath:
Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
L*W=352
2L+W=60
W=60-2L
L(60-2L)=352
60L-2L^2-352=0
-2L^2+60L-352=0
L^2-30L+176=0
(L-22)(L-8)=0
L-22=0
L=22 SOLUTION
L-8=0
L=8 SOLUTION
22*W=352
W=352/22
W=16 SOLUTION
PROOF
2*22+16=60
44+16=60
60=60

Answer by joyofmath(189) About Me  (Show Source):
You can put this solution on YOUR website!
Let the width of the pool = W and the length of the pool = L. We are told that the length is greater than W so L+%3E+W.
Assume the three sides that are fenced in are the two widths and one length, and that's with 60 yards of fencing. Thus,
(1) 2W%2BL+=+60
We also know that the area of the pool = 352. So,
(2) LW+=+352
Subtract 2W from both sides of (1) to get L+=+60-2W
Replace L with 60-2W in (2) to get
(3) %2860-2W%29W+=+352
Now, multiply 60-2W by W and we get 60W-2W%5E2.
So, (3) becomes 60W-2W%5E2+=+352 or
(4) 60W-2W%5E2-352+=+0
This is a quadratic equation which can also be written as -2W%5E2%2B60W-352+=+0.
We divide both sides by 2 and get -W%5E2%2B30W-176+=+0.
The coefficients of this quadratic equation are -1, 30, and -176.
Remember the quadratic equation, solving for w: W+=+%28-b+%2B-+sqrt%28b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
Replace a, b, and c with -1, 30, and -176 respectively and you get
W+=+%28-%2830%29+%2B-+sqrt%2830%5E2+-+4%2A%28-1%29%2A%28-176%29%29%29+%2F+%282%2A%28-1%29%29
Multiply things out and you get
W+=+%28-30+%2B-+sqrt%28900+-+704%29%29+%2F+-2 or
W+=+%28-30+%2B-+sqrt%28196%29%29+%2F+-2
sqrt%28196%29+=+14 so W+=+%28-30+%2B-+14%29+%2F+-2
So, W can equal %28-30+%2B+14%29+%2F+-2 which = 8
or W can equal %28-30+-+14+%29+%2F+-2 which = 22
From (1) we know that 2W%2BL+=+60. If W = 8 then 2%2AW%2BL+=+60 or 16%2BL+=+60 so L+=+44.
If W = 22 then 2%2AW%2BL+=+60 or 2%2A22+%2B+L+=+60 or 44%2BL+=+60 so L=16. But, L must be > W and 16 is not > 22 so W = 22 is not a solution.
So, L = 44. and from (1) we know that 2W%2BL+=+60 so 2W%2B44+=+60 so 2W+=+16 and W+=+8.
To verify our answer, that L+=+44 and W+=+8 we plug these values into (1) and (2):
2W%2BL+=+60 so 2%2A8+%2B+44 should = 60 and it does.
LW+=+352 so 44%2A8 should = 352 and it does.