quiz9-8.40-soln.

# quiz9-8.40-soln. - x cannot be bigger than 100(the total...

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MATH 101-B Fall 2004 Quiz 9 (8:40-10:30 Group) December 10, 2004 Time: 20 minutes Name: Student No: Follow the directions. No work No credit!! Problem 1 (2 pts.) Find the derivative of y cos sinh 3 x 2 ±± . Solution. Via chain rule, we have dy dx sin sinh 3 x 2 ±± cosh 3 x 2 ± 6 x . Problem 2 (8 pts.) If you are given a 100 feet of fencing and want to enclose a rectangular area up against a long straight wall, what is the largest area you can enclose? Solution. If we call the side lengths of the rectangle x and y , then the area function is given by A xy . Note that 100 feet will be used to fence only three sides, since one side is the wall. So, we also have the relation 100 2 x ± y (assuming the missing side (the wall) is y ). Solving this equation for y , we have y 100 2 x . Putting this into the area formula, we have A x ± x 100 2 x ± 2 x 2 ± 100 x . Note that since x represents a side length, it cannot be negative. Also,
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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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