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Practice Exam 3 Solved

# Practice Exam 3 Solved - Create assignment 57950 Exam 3 Nov...

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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. GAGAN’S 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: Let’s 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 0 ( 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

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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 < 0 , 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 0 = ( x 0 , y 0 ) is the point on the line 3 x + y = 5 closest to the origin, what is the value of x 0 ? 1. x 0 = 5 7 2. x 0 = 3 2 correct 3. x 0 = 15 11 4. x 0 = 5 4 5. x 0 = 15 22 Explanation: Geometrically, P 0 = ( x 0 , y 0 ) is the point P ( x, y ) on the line 3 x + y = 5 shown in P 5 5 3 (not drawn to scale) minimizing the length D = p x 2 + y 2 of the dashed line from P to the origin. Thus we have to minimize D ( x ) = £ x 2 + (5 - 3 x ) 2 / 1 / 2 on the interval £ 0 , 5 3 / . From the graph it is clear that this minimum value occurs at a
Create assignment, 57950, Exam 3, Nov 24 at 12:25 pm 3 point x 0 in ( 0 , 5 3 ) ; i.e. at a critical point of D .

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

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