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Unformatted text preview: 1 Practice exercises for Exam 3 (with answers) Just like the previous ones, this practice set was designed assuming you have solved all of the homework problems to date, and reviewed your lecture notes. Applied optimization 1. An indoor physical fitness room will consist of a rectangular region with a semicircle at each end (as shown on the picture). The perimeter of the room is to be a 200m track. Find the dimensions of the rectangular region that will maximize its area (i.e. the area of the rectangular region). Solution. Need to optimize the area A of the rectangle. Denote the radius of each semicircle by r and let h be the height of the rectangle. What is A ? It follows that the width of the rectangle is 2 r , hence A = 2 rh . The basic physical constraints are r > 0, h > 0. Now must account for the condition that the perimeter of the room is 200. Since each circular arc contributes half a circumference their lengths are πr . So the perimeter of the room is P = 2 h + 2 πr and we are told P = 200: 2 h + 2 πr = 200 Simplify the relation for the area by substituting h = 200- 2 πr 2 = 100- πr . Note that it follows r < 100 /π to ensure that h remains positive. So we need to maximize A , which becomes a function of r , and find its max on the interval < r < 100 /π . Answer: height = 50 m , width = 100 /π ≈ 31 . 83 m ( r = 50 /π ). 2. The width and the height of a box add up to 2 ft, and the length of the box is 1 ft more than its height. What dimensions will maximize the volume of this box? Solution. Let width be x , then height is 2- x , and the length is 1 + x . So the volume is V ( x ) = x (2- x )(1 + x ) =- x 3 + x 2 + 2 x with constraints 0 < x < 2. Note that we can actually include the end-points 0 ≤ x ≤ 2 (the box will simply collapse and have 0 volume then)....
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- Spring '08