Practice Exam 3 Solved - Create assignment, 57950, Exam 3,...

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Unformatted text preview: Create assignment, 57950, Exam 3, Nov 24 at 12:25 pm 1 This print-out should have 37 questions. Multiple-choice 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, wording-variable. 001 Stewart Example 5, page 282 Find the area of the largest rectangle that can be inscribed in a semi-circle 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 semi-circle 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 x-axis 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, wording-variable. 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, wording-variable. 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.

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Practice Exam 3 Solved - Create assignment, 57950, Exam 3,...

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