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Unformatted text preview: Create assignment, 57950, Exam 3, Nov 24 at 12:25 pm 1 This printout should have 37 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. GAGANS PRACTICE EXAM. CalC4g00Ex5 50:07, calculus3, multiple choice, > 1 min, wordingvariable. 001 Stewart Example 5, page 282 Find the area of the largest rectangle that can be inscribed in a semicircle of radius 2. 1. Area = 4 sq. units correct 2. Area = 3 sq. units 3. Area = 1 sq. units 4. Area = 0 sq. units 5. Area = 2 sq. units Explanation: Lets take the semicircle to be the upper half of the circle x 2 + y 2 = 4 having radius 2 and center at the origin. For the rectangle we take one side on the xaxis and one corner at a point P ( x, y ) on the semi circle as shown in P ( x,y ) 2 2 Then the area of the rectangle is given by A ( x ) = 2 xy = 2 x p 4 x 2 . We have to maximize A ( x ) on the interval [0 , 2]. Now A ( x ) = p 4 x 2 x 2 4 x 2 = 4 2 x 2 4 x 2 . Thus the critical points of A ( x ) occur at x = 2 2 , 2 2 , only one of which lies in [0 , 2]. But A (0) = 0 , A 2 2 = 4 , A (2) = 0 . Consequently, max. area = 4 sq. units . CalC4g11a 50:07, calculus3, multiple choice, > 1 min, wordingvariable. 002 The canvas wind shelter x is to be constructed for use on Padre Island beaches. It is to have a back, two square sides, and a top. If 600 square feet of canvas are available, find the length x of the shelter for which the space inside is maximized assuming all the canvas is used. 1. length = 21 feet 2. length = 19 feet Create assignment, 57950, Exam 3, Nov 24 at 12:25 pm 2 3. length = 22 feet 4. length = 20 feet correct 5. none of these Explanation: Let y be the width of the shelter. Then the ends of the shelter are squares of side length y , so the volume V of the shelter is given by ( ) V ( x,y ) = xy 2 , while its area is given by ( ) 2 xy + 2 y 2 = 600 . Using ( ) to eliminate x from ( ) we see that V ( y ) = y 2 600 2 y 2 = 300 y y 3 . The maximum volume will occur at a critical point of V , i.e. , when dV dy = 300 3 y 2 = 0 . But then y = 10. Now d 2 V dy 2 fl fl fl y =10 = 60 < , so the maximum volume occurs when y = 10. From ( ) it thus follows that the maximum volume occurs when length = 20 feet . Notice that the negative solution y = 10 does not have a practical meaning, but from a strictly mathematical point of view it corre sponds to a minimum value of V . CalC4g15s 50:07, calculus3, multiple choice, > 1 min, wordingvariable. 003 If P = ( x , y ) is the point on the line 3 x + y = 5 closest to the origin, what is the value of x ?...
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This note was uploaded on 02/28/2010 for the course M 56200 taught by Professor Radin during the Fall '09 term at University of Texas at Austin.
 Fall '09
 RAdin

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