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Approximate the length and width of a rectangle whose perimeter is 96 feet and whose area is 500 square feet accurate to three decimal places?
Rnachbar15
Perimeter, P = 96 ft.
Area, A = 500 ft^2.
A = LW
500= LW
L = 500/W
P = 2L + 2W
96 = 2L + 2W substitute 500/W for L;
96 = 2(500/W) + 2W divide by 2 to both sides;
48 = 500/W + W multiply by W to both sides;
48W = 500 + W^2
0 = 500 + W^2 - 48W or,
W^2 - 48W + 500 = 0
W = [48 +,- square root of [48^2 - (4)(1)(500)]/2
W = [48 +,- square root of (2304 -2000)]/2
W = (48 +,- square root of 304)/2
W = (48 + 17.436)/2 or W = (48 - 17.436)/2
W = 65.436/2 or W = 30.564/2
W = 32.718 ft or W = 15. 282 ft
L = 500/W = 500/32.718 or L = 500/15.282
L = 15.282 ft or L = 32.718 ft
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