MCB4UW Optimization Problems Handout 4.6

Transcription

1 MCB4UW Optimization Problems Handout A rectangular field along a straight river is to be divided into smaller fields by one fence parallel to the river and 4 fences perpendicular to the river. Find the maximum area that can be enclosed if 1600m of fencing is available.. Find the area of the largest rectangle that can be inscribed inside the ellipse x y A box with an open top is to be made from a square piece of cardboard, of side length 100 cm, by cutting a square from each corner and then folding up the sides. Find the dimensions of the box of largest volume. 4. An experimental farm has 600 m of fencing with which to enclose and subdivide a rectangular field into 4 equal plots of land. a) What is the largest area that can be enclosed? b) What is the largest area that can be enclosed if each side of each plot is required to be at least 50 m long? 5. A soft drink can in the shape of a right circular cylinder is to have a capacity of 50 mm. If the diameter of the can must be no less than 4 cm and no greater than 8 cm, find the dimensions of the can that will use the least amount of material (include top, bottom, and side). What is the ratio of height to diameter for this can? 6. If the sum of two non-negative numbers is 10, how should the numbers be chosen so that the sum of their squares is a) a maximum? b) a minimum?

2 7. In microcomputers most of the components are squeezed into a single box-shaped block. If the block has a length equal to twice the width and if the total surface area of the block must be 00 cm in order to dissipate the heat produced, find the dimensions for the maximum volume of the block. 8. The perimeter of an isosceles triangle is 6 cm. Find the length of the sides of the triangle of maximum area. 9. A 400 m track has the shape of two semi-circles at the ends of a rectangle. The straight sections of the track must be at least 100 m in length, and the radius of the semi-circles must be at least 0 m. Find the dimensions of the track that encloses a) the maximum area b) the minimum area 10. A car rental agency has 00 cars. The owner finds that at a price of $6 per day he can rent all the cars. For each $ increase in price, the demand is less and 5 fewer cars are rented. What price will maximize the total revenue? 11. A piece of wire 180 cm long is cut into 6 sections, of one length and 4 of another length. Each of the two sections having the same length is bent into the form of a circle, and the two circles are then joined by the 4 remaining sections to make a frame for a model of a right circular cylinder. Find the lengths of the sections that will maximize the volume of the cylinder.

3 1. The current ticket price at a local theatre is $4, and the theatre attracts an average of 50 customers per show. Every $0.0 increase in ticket price reduces the average attendance by 10 customers, while every $0.0 decrease results in 10 extra customers. a) Let x represent the change in ticket price in dollars, Show that the revenue R from ticket sales depends on x according to R x x x. b) If the seating capacity is 400, show that x 5. c) Find the ticket price that will maximize revenue. 1. Bob and Sue are both training for a marathon. Bob s house is located 0 km north of Sue s house. At 9:00 on Saturday morning, Bob leaves his house and jogs south at 8 km/h. At the same time, Sue leaves her house and jogs east at 6 km/h. When are Bob and Sue closest together, given that they both run of.5 h? 14. The cost of laying a power line underwater is k (>1) times that of underground. An island is m km from the shore and a power station is a distance L km from the shore which is closest to the island. How should the power line be laid so that the cost is minimum when going from the power station to the island.. m L p.s. island, ) y ( x ) 15. Determine the minimal distance the point ( to the curve given by. 16. A paper drinking cup in the form of a right circular cone can be made from a circular piece of waxed paper by removing a sector and joining the edges OA and OB. Show that the maximum capacity of a cup that can be formed from a disc of radius R π R centimetres is V c m. 9 O R A B

4 17. Find the area A of the largest rectangle that can be inscribed inside the ellipse x a y + b A sheet of paper 8 cm by 10 cm is folded as in the diagram. How should the sheet be folded so as to minimize the length AB of the fold. C B 8 10 A

5 Handout 4.6 Solutions 1. Label: x x x x Let x represent the vertical fence in m Let y represent the horizontal fence in m Let A represent the area in m Given: 1600 m of fence Required: To maximize A Relationship: A xy 4x+ y 1600 y Work: Since we want to maximize Axy, we must express A in terms of one variable. Set y x Substitute this into Axy A x y b x x 1600x 4x g b g 0 We need to find the extreema of A(x), 0 x 400, by solving for x in A x b g A x x 0 x 00 x We need to check, the endpoint of the restrictions, and x00 b g b b A 0 0 g g A A Conclusion: The maximum area is m

6 . Label: Given: Let L represent the length of the rectangle Let W represent the width of the rectangle Let A represent the area of the rectangle y x y Required: Relationship: To maximize A L x W y A L W Work: Since we want to maximize ALW, we must express A in terms of one variable. x y A L W bxgbyg 4x + 9y 6 4xy 4 y c x 9 9 h y ± x 9 F I This gives us A 4x x HG K J 9, since area is positive need only positive y term We need to find the extreema of A(x), 0 x, by solving for x in A x bg F I HG K J L + b g O NM c h b g QP A x 4 1 x x x x x x 9 x x, need only positive x value We need to check, the endpoint of the restrictions, and x00 b g 0

7 bg F A 0 0 I A 1 HG K J A 0 bg Conclusion: The maximum area is 1.

8 . Label: Let L represent the length of the box in cm Let W represent the width of the box in cm Let h represent the height of the box in cm Let x represent the length of cut out piece in cm Let V represent the volume of the box in cm Given: Side of cardboard is 100cm Required: To maximize V Relationship: V L W h L 100 x W 100 x h x Work: Since we want to maximize V, we must express V in terms of one variable. V L W h b gb gbxg 100 x 100 x 4x 400x x b g 0 We need to find the extreema of V(x), 0 x 50, by solving for x in V x V x 1x 800x x 800x ( x )( x )

9 Therefore x50, and x 50 We need to check, these points and the endpoint of the restrictions A 0 0 A b g F 50I HG K J bg A 50 0 Conclusion: Therefore dimensions are L cm W cm h 50 c m

10 4a. Label: Let x represent the vertical fence in m Let y represent the horizontal fence in m Let A represent the area in m Given: 600 m of fence Required: To maximize A Relationship: A xy 5x+ y 600 Work: Since we want to maximize Axy, we must express A in terms of one variable. b 5 Set y x 70 Substitute this into Axy A x y b 5 x x x 5 x g g We need to find the extreema of A(x), b g 0 0 x 70, by solving for x in A x b g A x x x 0 x 60 We need to check, the endpoint of the restrictions, and x60

11 b g b g b g A 0 0 A A 70 0 Conclusion: The maximum area is 4000m 4b. Exactly the same as 4a, except the restriction now is 50 x 0 with 5x+ 8y 600 This gives us the following values for A b g A Conclusion: The maximum area is 0000 m

12 5. Label: Let r represent the radius of the can in mm Let d represent the diameter of the can in mm Let h represent the height of the can in mm Let A represent the area of the can in mm Let V represent the volume of the can in mm Given: V 50mm r 4 Required: d, h, d:h when A is a minimum Relationship: V πr h A πr + πrh Work: Since we want to maximize A, we must express A in terms of one variable say r. 50 πrh 50 h πr Substitute this into A πr + πrh 50 A πr πr π r 500 πr + r + F H G I K J We need to find the extreema of A(r), b g 0 r 4, by solving for x in A r

13 A bg r 4πr 500r 4πr 500r 0 4πr r π r 419. We need to check this and the endpoint of the restrictions b g b bg A 77 A g A 4 6 Now let's find h 50 h πr 50 π 419. b d g (.419) Conclusion: The dimensions are d6.879 mm and h6.879 mm, with a ration of d:h1:1

14 6. Label: Let x represent the first number Let y represent the second number Let S represent sum of their squares Given: x+ y 10 Required: x, y when S is a maximum Relationship: x + y Work: S Since we want to maximize S, we must express S in terms of one variable say x. x+ y 10 x + y S Substitute y 10 x into x + y S b S x + 10 x x x+ x x 0x+ 100 g We need to find the extreema of Sx b g, 0 x 10, by solving for x in S x b g 0 4x 0 0 x 5 We therefore need to check S b g b g bg S 5 50 S Therefore the max occurs when both numbers are 5 b) Therefore the min occurs when one number is 0 and the other is 10

15 7. Label: Let L represent the length in cm Let W represent the width in cm Let h represent the height in cm Let S represent the surface area in cm Let V represent the volume in cm Given: S 00cm Required: L, W, h, when V is a maximum Relationship: L W S LW + Lh+ hw V LWh Work: Since we want to maximize V, we must express V in terms of one variable say W. 00 LW + Lh + Wh 100 LW + Lh + Wh Substitute LW into this b g 100 WW+ Wh+ Wh W + Wh 100 W h W b g Let's substitute everything into V

16 V L Wh b g WW F HG 100 W W 00W 4W W 00 4 W W 4 I KJ b g 0 We need to find the extreema of V(W), by solving for x in V W only restriction 00 bg b g V W 00 4W V W 0 4W 0 4W W W , W0 is We don't need to check this, as restriction produces a volume of zero

17 L W F H G I 5 6 K J W h W F I 5 6 HG F H G I K J KJ Conclusion: The dimensions are cm.

18 8. Label: Let x represent the equal sides of the triangle in cm Let y represent other side of the triangle in cm Let h represent the height of the triangle in cm Let P represent perimeter in cm Let A represent the area in cm Given: P 6 Required: x, y when A is a maximum Relationship: P x + y 1 A x Work: y yh F H G I K J + h Since we want to maximize A, we must express A in terms of one variable. Set y 6 x F y Set h x H G I K J b x 18 x 6 x 9 g Substitute this into A 1 A b x x 6 g 6 9 b g 6 x x 9

19 b g 0 We need to find the extreema of A(x), 0 x 18, by solving for x in A x 6 9 ( 18 )( 9) ( x) 0 A x x + x x A ( x ) 1 6 x x x x 6 x 9 x x 9x 108 x 1 We need to check this and the endpoint of the restrictions b g b g b g A 0 0 A 1 6 S 18 0 Now let's find y. bg bg d 6 x Conclusion: The triangle is an equilateral triangle with all sides 1cm

20 9. Label: Let A represent the area in m Let P represent the perimeter in m Let r represent the radius in m Let d represent the straight in m Given: P400m Required: A when A is a maximum and a minimum Relationship: A πr + rd P πr+ d Work: Since we want to maximize A, we must express A in terms of one variable say r. 400 πr+ d d 00 πr Substitute this into Area formula A πr + r 00 πr πr + 400r πr b r πr, 0 r π g Let's find the derivative and set it equal to zero and solve for r da 400 πr dr πr 00 r π Since this is outside the restrictions, we need only check the end points

21 A π A b g F I HG K J π π Conclusion: This minimum area occurs when r0m the maximum area when r 100 π

22 10. Label: Let x represent the number of $ increases Let C represent the number of cars rented Let P represent the rental price Let R represent the revenue Given: C00-5x P6+x Required: P when R is a maximum Relationship: R C P Work: Since we want to maximize R, we must express R in terms of one variable say x. R C P b gb g 00 5x 6 + x 10x + 0x+ 700, 0 x 40 Let's find the derivative and set it equal to zero and solve for x dr 0x + 0 dx 0 0x + 0 x 11 b g b g bg R R R 40 0 Conclusion: This maximum occurs when x11 or rental price is $6+$$58

23 11. Label: Let C represent the circular wire in cm Let L represent the straight wire in cm Let V represent the volume of the cylinder in cm Let d represent the diametre of circle in cm. Let r represent the radius of the circle in cm. Let A represent the Area of a circle in cm Given: C+4L180 Required: L and C when volume is a maximum. Relationship: Work: C π d A πr V AL V AL πr L C C L, r 4π π Must write V in terms of one variable 4L 180 C L Conclusion: 90 C Therefore V dv dc C 4 π F HG 90 C I K J 180C C 8π We cant C when derivative equals zero, that is when C0 or C60 Since C60 is valid in domain then the circular piece is 60cm long and the straight piece is 15cm long

24 1a) Revenue equals the product of price and number of tickets sold. Letting x represent the 5 times the number of $0.0 increases, the price is then (x+4). The number of seat sold is then ( 50 50x) This gives us the a revenue of ( x )( 1b) Minimum capacity is x 50x 50 x 5 Maximum capacity is x x x Therefore x 5 1c) R( x) ( x+ 4)( 50 50x) R ( x) []( x) + ( x+ 4)[ 50] x x x 1 Check critical points R R R 5 0 Therefore the ticket price is $ x )

25 1. If Bob starts at I, he reaches point J after time t hours. Then IJ8t km and JA 0 8t km. If Sue starts at point A, she reaches point B after t hours and AB6t km. Now the distance they are apart is s Jb ( JA) ( AB) + ( 0 8t) ( 6t) + t t+ t Take derivative to find minimum () s t t t+ t 0 100t t 0t t t 0t t t 1.6 Check for extreme values. s s s Therefore, the minimum value of s(t) is 1km and it occurs at 10:6.

26 14. Let x represent the distance along the shore where the power line reaches land which is closest to the island. Let C represent the cost m x L-x p.s. island The cost is made up of land cost of L x and water cost k m + x C x k m + x + L x 0 x L Set the derivative to zero, and solve for x to find critical points. C ( x) 0 kx m + x kx 1 1 m + x x k m 1 This is the distance that produces the lowest cost.

27 15. Let the point on the curve be ( x, x ). Therefore the distance between this point and (, ) is ( x ) ( x ) ( ) d x + x ( x ) ( x ) ( x 1)( x 16x ) d x+ + x x + 0 x+ + x x + 0 x+ + x x x x + x + This gives us x1 as a critical point. d The minimal distance is

28 16. Label: Let R represent the radius of the circle in cm Let V represent the volume in cubic cm Let r represent the radius of cone in cm Let h represent the height of cone in cm Relationships: 1 V π r h R h + r Remember R is a constant. Work: h R r 1 V π r R r 1 dv 1 1 πr R r πr ( R r ) ( r) dr + ( ) π r R r R r Setting to zero and solving for r π r R r 0 πr πrr r R. ;only positive required therefore R h R R

29 1 This gives us V π r h R R 1 π π R 9

30 x 17. Given a y b +, then solving for y we have y a x b a 1 We want the maximum area where A 4xy Therefore b A 4x a a x ( x ) 4b a A a a x Setting the derivative to zero and solving for x, we have ( x ) 4b a 0 x a a x This gives us an area of a b a A 4 a a ab The area of an ellipse is A π ab Therefore the ratio is π ab : ab π :

31 18. Label: d C a b B 8 c 10 A Let a represent BC in cm Let b represent AC in cm Let c represent AB in cm Let d represent the distance between C and upper left hand corner Given: the dimensions are 10 cm 8 cm Required: d, a, and b when c is a minimum. Relationships: a d + ( 8 a) b 8 + ( b d) c a + b Work: We need to write c in terms of one variable (look at the relationships and you should notice that the variable will be d). ( 8 ) a d + a a d a+ a 16a d + 64 d + 64 a 16 b 8 + b d b 64 + b bd + d + bd d 64 d + 64 b d () 1 We now sub in (1) and () in c a + b

32 c a + b d + 64 d d c ( + 64) + 64( + 64) d d d ( d + 64) 56d ( d + 64) 56d, 4 d 8 16d The end point value of 4 comes from substituting the max value of b10 into () Let s take the derivative to find critical points. c d 0 ( d + 64 ) d ( 16d) ( d + 64) [ 16] ( d + 64) 48d 16( d + 64) ( d + 64) ( d ) ( d ) ( d ) 0 56d 1 56d 56d d d 4 Let s check all extreme to find the minimum value. c cm c cm c cm Therefore AB has a minimum value 6 cm occurs when d 4 and b 6 cm cm, a6 cm,