Question 1013245
 the perimiter of rectangle ABCD is 20 inches. find the least value, in inches, of diagonal AC. answers must be in exact form. 
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P = 2(L + W)
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20 = 2(L+W)
L+W = 10
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Diam = sqrt[L^2 + W^2)
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Substitute for "L" to get:
D = sqrt[(10-W)^2 + W^2]
D = sqrt[2W^2 - 20W + 100]
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Take the derivative to get:
D' = (1/2)(4W-20)/sqrt(2W^2-20W+100)
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Solve:: 2W-10 = 0
W = 5
Then L = 5
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Ans: D = sqrt[5^2 + 5^2] = sqrt(2*25) = 5sqrt(2)
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Cheers,
Stan H.
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