word problem?

Question

An architect has designed a motel pool within a rectangluar area that is fenced on three sides. If she uses 60 yards of fencing to enclose an area of 352 square yards, then what are the dimensions (L) and (W)? Assume (L) is greater than (W).

Answer 1

Let the length of the area be x and the width y yard respectively
She may enclose the area in 2 ways i.e 2x+y or x+2y
If the fencing is in the first pattern,i.e.2x+y,then according yo the problem
2x+y=60.......
or y=60-2x.......(1)
xy=352
orx(60-2x)=352
or,60x-2x^2=352
or -2x^2+60x-352=0
or,x^2-30x+176=0 [dividing both sides by -2]
or,(x-22)(x-8)=0
or x=22 or 8
If x=22,y=60-2*22=60-44=16
If x=8,y=60-2*8=60-16=44
rejecting the second set of values becausethe width is more than the length,the length of the area is 22yards and the width is 16 yards

Answer 2

let y the length that is opposite of the pool and x be the the other two sides

P = x + x + y
60 = 2x + y
y = 60 - 2x

Area = xy
352 = x(60 - 2x)
352 = 60x - 2x^2
0 = -2x^2 + 60x - 352

use quadratic formula and you'll get x = 8 and 22

y = 60 - 2x
y = 60 - 2(8)
y = 44

y = 60 - 2(22)
y = 16


so the demensions are 16yrd x 22yrd OR 8yrd x 44yrd

Answer 3

A=L*W=352,
2*L+W=60
W=60-(2*L)
Substitute,
352=L*(60-(2L))
0=-2L^2 + 60L -352
0=2L^2 - 60L +352
Use quadratic formula to solve for L
L=(1/4)*(60 +/- sqrt(3600-2816))
L=(1/4)*(60 +/- 28)
L=22, L=8
But A=L*W=352, and L>W
So if L=22, W=16
and if L=8, W=44
therefore, L=22, W=16

Answer 4

L + 2W = 60 yds so L = 60 - 2W
LxW = 352 yds^2
(60-2W)(W) = 352
60W - W^2 =352
W^2 -60W-352=0
Solve using the quardatic equation formula