Unformatted text preview: x cannot be bigger than 100 (the total fencing material). So, our task is to maximize (find the global max.) the function A & x ± on the interval x ± ² 0,100 ³ . A ² & x ± & & 4 x ± 100 Since the derivative is defined for any x , the only critical points comes from A ² & x ± & & 4 x ± 100 & 0. So, x & 25 is the unique critical point. Note that this is a local max. point by the second derivative test, since A ²² & x ± ³ 0. Since there is no other critical point and the end points of the interval clearly yield 0 area, the global max. is attained when x & 25. So, with 100 feet of material, the largest such area one can fence is A & 25 ± & 1250 square feet....
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 Spring '10
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 Math, Calculus, Chain Rule, Derivative, MATH 101B Fall

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