Problema Solution
Find the length and the width of a rectangle whose perimeter is 48ft and whose area is 140 square feet.
Answer provided by our tutors
let
l = the length of the rectangle, l>0
w = the width of the rectangle, w>0
l > w
the perimeter of the rectangle P = 2(l + w)
2(l + w) = 48 divide both sides by 2
l + w = 24
the are of the rectangle A = l*w
l*w = 140 => l = 140/w
if we plug the value for l in l + w = 24 we get
140/w + w = 24
by solving the quadratic equation we find 2 solutions
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w = 10ft => l = 24 - w = 24 - 10 = 14 ft
w = 14ft => l = 24 - w = 24 - 14 = 10 ft but in this case l<w (10<14) thus the dimensions of the rectangle are
w = 10 ft
l = 14 ft
the length of the rectangle is 14 ft.
the width of the rectangle is 10 ft.