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?

Answer 1

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