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quiz9-10.40-soln.

# quiz9-10.40-soln. - in the domain of D ± x ² So the...

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MATH 101-B Fall 2004 Quiz 9 (10:40-12:30 Group) December 10, 2004 Time: 20 minutes Name: Student No: Follow the directions. No work ° No credit!! Problem 1 (2 pts.) Verify the identity cosh 2 x ° sinh 2 x ° 1. Solution. Recall the definitions of cosh x and sinh x : cosh x ° e x ± e ° x 2 sinh x ° e x ° e ° x 2 . So, cosh 2 x ° sinh 2 x ° e x ± e ° x 2 2 ° e x ° e ° x 2 2 ° 1 4 °± e x ± e ° x ² 2 ° ± e x ° e ° x ² 2 ³ ° 1 4 °± e 2 x ± 2 ± e ° 2 x ² ° ± e 2 x ° 2 ± e ° 2 x ²³ ° 1. Problem 2 (8 pts.) Of all the rectangles with given area A , which has the shortest diagonal? Solution. If we call the side lengths of the rectangle x and y , then the square of the diagonal is x 2 ± y 2 . Since the area is A , we have the relation xy ° A . Solving this equation for y , we have y ° A x . Substituting the result in the diagonal, it becomes D ± x ² ° x 2 ± A 2 x 2 ° x 4 ± A 2 x 2 . Note that x is necessarily positive. So, our task is to minimize (find the gloabl min.) of the function D ± x ² on the interval x ± ± 0, ² ² . D ² ± X ² ° 4 x 5 ° ± x 4 ± A 2 ² 2 x x 4 ° . 2 x 5 ° 2 A 2 x x 4 ° 2 ± x 4 ° A 2 ² x 3 ° 2 ± x 2 ° A ²± x 2 ± A ² x 3 The point x ° 0 makes the derivative undefined but it is not a critical point as it does not lie
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Unformatted text preview: in the domain of D ± x ² . So, the critical point(s) will come from the zeros of the derivative function. Since A ³ 0 (it represents area), the only zeros of D ² ± X ² are x & ´ A . Only x & A lies in the interval we are interested in, so it is the only critical point for us. Looking at the derivative function, we see that it has positive values on the interval ± 0, A ² , and negative values on ± A , ² ² . So, D ± x ² decreases on ± 0, A ² and increases on ± A , ² ² , creating a local and global min. at x & A . This means that the rectangle of area A with the largest diagonal arises if the side lengths are x & A , and y & A x & A , i.e. it is a square....
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