Question 981235
Let l be the length and w be the width
Perimeter (P) = 2l + 2w
Area (A) = l * w
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A. let l be length and w be width
B. P = 2l + 2w, A = l * w
C. we have two equations in two unknowns, solve by substitution
D. we are given
1) 86 = 2l + 2w
2) 450 = l * w
consider the first equation
86 = 2l + 2w
divide both sides of = by 2
43 = l + w, then
l = 43 - w
now substitute for l in our second equation
450 = (43 - w) * w
450 = 43w - w^2
rearrange terms
w^2 - 43w + 450 = 0
factor the quadratic equation
(w - 25) * (w - 18) = 0 and
w = 25 or w = 18
note that 25 * 18 = 450 = A, therefore
choose l = 25 m and w = 18 m
E.check our dimensions with the equation for P
86 = (2 * 25) + (2 * 18)
86 = 50 + 36
86 = 86
our dimensions check
F.if we double our dimensions we have l = 50 and w = 36 then
P' = 2*50 + 2*36 = 172 therefore we double P
A' = 50 * 36 = 1800 therefore we quad-rouble (4 times) A