HW 11 Gagan Solutions

# HW 11 Gagan Solutions - Gauss, Karl Friedric Homework 11 R...

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Gauss, Karl Friedric – Homework 11 – R Gompf – 57965 – Oct 28, 2004 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – ±nd all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Stewart Example 5, page 282 Find the area of the largest rectangle that can be inscribed in a semi-circle of radius 6. 1. Area = 36 sq. units correct 2. Area = 39 sq. units 3. Area = 37 sq. units 4. Area = 38 sq. units 5. Area = 40 sq. units Explanation: Let’s take the semi-circle to be the upper half of the circle x 2 + y 2 =3 6 having radius 6 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 ) 6 6 Then the area of the rectangle is given by A ( x )=2 xy =2 x p 36 x 2 . We have to maximize A ( x ) on the interval [0 , 6]. Now A 0 ( x )= p 36 x 2 x 2 36 x 2 = 36 2 x 2 36 x 2 . Thus the critical points of A ( x ) occur at x = 6 2 , 6 2 , only one of which lies in [0 , 6]. But A (0) = 0 ,A ³ 6 2 ´ =36 (6) = 0 . Consequently, max. area = 36 sq. units . 002 (part 1 of 1) 10 points Find the positive number such that the sum of 6 times this number and 5 times its reciprocal is as small as possible. Correct answer: 0 . 912871 . Explanation: Let x be a positive number. Then we have to ±nd which x 0 in (0 , ) minimizes the func- tion f ( x )=6 x + 5 x . Now f 0 ( x 5 x 2 ,f 0 ( x 10 x 3 . Thus f is concave up everywhere on (0 , ) and x 0 = r 5 6 is the critical point of f for which f ( x 0 )i s the absolute minimum value of f on (0 , ). Consequently, x 0 = r 5 6 =0 . 912871 0 . 912871 . 003 (part 1 of 1) 10 points

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Gauss, Karl Friedric – Homework 11 – R Gompf – 57965 – Oct 28, 2004 2 A2 4 0 × 24 0 square sheet of metal is made into an open box by cutting out a square at each corner and then folding up the four sides. Determine the maximum volume, V max ,ofthe box. 1. V max = 1044 cu. ins. 2. V max = 1039 cu. ins. 3. V max = 1029 cu. ins. 4. V max = 1034 cu. ins. 5. V max = 1024 cu. ins. correct Explanation: Let x to be the length of the side of the squares cut from each edge. Then the volume of the resulting box is given by V ( x )= x (24 2 x ) 2 . Di±erentiating V with respect to x we see that dV dx =(24 2 x ) 2 4 x (24 2 x ) . The critical points of V are thus the solutions of 3 x 2 48 x + 144 = 0 , i.e. , x 1 =4 ,x 2 =12 , where the second one can be disregarded for practical reasons. At x = x 1 , therefore, V ( x ) becomes V max = 1024 cu. ins. . 004 (part 1 of 1) 10 points A homeowner wants to build a fence to enclose a 80 square yard rectangular area in his backyard. Along one side the fence is to be made of heavy-duty material costing \$9 per yard, while the material along the remaining three sides costs \$1 per yard. Determine the least cost to the homeowner. 1. least cost = \$70 2. least cost = \$80 correct 3. least cost = \$85 4. least cost = \$65 5. least cost = \$75 Explanation: Let x be the length of the side made of the heavy-duty material and y theleng tho fan adjacent side. Then we want to minimize the cost function C ( x, y )=1 0 x +2 y, subject to the constraints xy =8 0 , y > 0 .
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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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HW 11 Gagan Solutions - Gauss, Karl Friedric Homework 11 R...

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