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Unformatted text preview: f ( x ) = 2 x + 3000 x , x ∈ (0 , ∞ ) , where f ( x ) = 23000 x 2 = 2 x 23000 x 2 = ( , at x = ± √ 1500 = ± 10 √ 15 , undeﬁned , at x = 0 . But only the point x = 10 √ 15 is in the given domain of f , so this is the only critical point of f . Using f 00 ( x ) = 6000 x 3 , it follows that f 00 (10 √ 15) = 6000 10 3 15 3 / 2 > , which conﬁrms that the function f has a local minimum at x = 10 √ 15 from the Second Derivative Test. Note also that f 00 ( x ) > , for all x ∈ (0 , ∞ ) , so that the graph of f ( x ) is concave up over its entire domain. From this it follows that the local minimum is an absolute minimum over this domain. Thus the optimal value of x is x = 10 √ 15 and the corresponding value of y is y = 3000 10 √ 15 = 300 · √ 15 15 = 20 √ 15 . The pen using the least amount of fencing has dimensions 10 √ 15 ft × 20 √ 15 ft . 2...
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 Fall '10
 KIHYUNHYUN
 Calculus, Critical Point, Derivative, Optimization

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