This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Tran, Tony – Homework 12 – Due: Nov 13 2007, 3:00 am – Inst: Samuels 1 This printout should have 17 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A rectangle is inscribed in a semicircle of diameter d as shown in d What is the largest area the rectangle can have when d = 6? 1. Area = 12 sq. units 2. Area = 9 sq. units correct 3. Area = 11 sq. units 4. Area = 13 sq. units 5. Area = 10 sq. units Explanation: Let’s take the semicircle to be the upper half of the circle x 2 + y 2 = 9 having radius 3 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 ) 3 3 Then the area of the rectangle is given by A ( x ) = 2 xy = 2 x p 9 x 2 . We have to maximize A ( x ) on the interval [0 , 3]. Now A ( x ) = 2 p 9 x 2 2 x 2 √ 9 x 2 = 2(9 2 x 2 ) √ 9 x 2 . Thus the critical points of A ( x ) occur at x = 3 √ 2 , 3 √ 2 , only one of which lies in [0 , 3]. But A (0) = 0 , A ‡ 3 √ 2 · = 9 , A (3) = 0 . Consequently, max. area = 9 sq. units . keywords: optimization, semicircle, critical point, maximum area, area rectangle 002 (part 1 of 1) 10 points A 6 00 × 6 00 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 , of the box. 1. V max = 36 cu. ins. 2. V max = 16 cu. ins. correct 3. V max = 26 cu. ins. 4. V max = 31 cu. ins. 5. V max = 21 cu. ins. Tran, Tony – Homework 12 – Due: Nov 13 2007, 3:00 am – Inst: Samuels 2 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 (6 2 x ) 2 . Differentiating V with respect to x we see that dV dx = (6 2 x ) 2 4 x (6 2 x ) . The critical points of V are thus the solutions of 3 x 2 12 x + 9 = 0 , i.e. , x 1 = 1 , x 2 = 3 , where the second one can be disregarded for practical reasons. At x = x 1 , therefore, V ( x ) becomes V max = 16 cu. ins. . keywords: optimization, box, maximum, vol ume, constraint 003 (part 1 of 1) 10 points A homeowner wants to build a fence to enclose a 180 square yard rectangular area in his backyard. Along one side the fence is to be made of heavyduty 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 = $120 correct 2. least cost = $130 3. least cost = $115 4. least cost = $125 5. least cost = $110 Explanation: Let x be the length of the side made of the heavyduty material and y the length of an adjacent side. Then we want to minimize the cost function C ( x,y ) = 10 x + 2 y , subject to the constraints xy = 180 , x, y > ....
View
Full Document
 Fall '08
 schultz
 Inverse function, Tran, Samuels, cu. ins

Click to edit the document details