Question 253415
If the perimeter of a rectangle is 8√2 cm, what is the smallest possible value of the length of one of its diagonals in cm? 
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Perimeter = 2(L + W) = 8sqrt(2) cm
L+W = 4sqrt(2) cm
Note:
W = 4sqrt(2)-L
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Diagonal = sqrt(L^2 + W^2)
Substitute for W:
D(L) = sqrt(L^2 + (4sqrt(2)-L)^2)
D(L) = sqrt(L^2 + [L^2 - 8sqrt(2)L +32]
D(L) = sqrt(2L^2 -8sqrt(2)L + 32)
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You have a quadratic with a = 2 ; b = -8sqrt(2)
Minimum occurs when L = -b/2a = 8sqrt(2)/(4) = 2sqrt(2)
W = 4sqrt(2)-2sqrt(2) = 2sqrt(2)
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Minimum Diagonal:
D^2 = L^2 + W^2
D^2 = (2sqrt(2))^2 + (2sqrt(2))^2
D^2 = 8 + 8
D = 4 cm
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Cheers,
Stan H.