612 MHR Principles of Mathematics 9 Solutions. Optimizing Measurements. Chapter 9 Get Ready. Chapter 9 Get Ready Question 1 Page 476.

Transcription

1 Chapte 9 Optimizing Measuements Chapte 9 Get Ready Chapte 9 Get Ready Question Page 476 a) P = w+ l = = = 60 A= lw = 0 0 = 00 The peimete is 60 cm, and the aea is 00 cm. b) P = w+ l = = = 8 A= lw =. 5.8 = The peimete is 8 m, and the aea is m. 6 MHR Pinciples of Mathematics 9 Solutions

2 Chapte 9 Get Ready Question Page 476 a) C = π = π 4 5. A= π = π The cicumfeence is appoximately 5. cm, and the aea is appoximately 50. cm. b) C = π = π A= π = π The cicumfeence is appoximately.8 cm, and the aea is appoximately. cm. MHR Pinciples of Mathematics 9 Solutions 6

3 Chapte 9 Get Ready Question Page 476 a) V = lwh = = 0 SA = A + A + A bottom sides font ( ) ( ) ( = = = 04 The volume is 0 cm, and the suface aea is 04 cm. ) b) V = lwh = = 4.9 SA = A + A + A bottom sides font ( ) ( ) ( = = = 4.54 The volume is 4.9 m, and the suface aea is 4.54 m. ) 64 MHR Pinciples of Mathematics 9 Solutions

4 Chapte 9 Get Ready Question 4 Page 476 a) V = π h = π SA = π + πh = π 7 + π 7 86 The volume is appoximately 847 cm, and the suface aea is appoximately 86 cm. b) V = π h = π SA = π + πh = π.5 + π The volume is appoximately 4 m, and the suface aea is appoximately 9 m. MHR Pinciples of Mathematics 9 Solutions 65

5 Chapte 9 Get Ready Question 5 Page 477 a) V = lwh = = 07 SA = A + A sides bottom ( ( 8 48) ( 8 8) ) ( 8 48) = + + = = 80 The volume is 07 cm, and the suface aea is 80 cm. V = 6 6 = 07 SA = A + A sides bottom ( ( 6 6) ( 6) ) ( 6 ) ( 5 84) 9 = + + = + + = 088 The volume is 07 cm, and the suface aea is 088 cm. b) The volumes of the two boxes ae equal. c) The second containe equies less mateial. 66 MHR Pinciples of Mathematics 9 Solutions

6 Chapte 9 Get Ready Question 6 Page 477 a) V = π h = π SA = π + πh = π 0 + π The volume is appoximately 5 m, and the suface aea is appoximately 87 m. V = π h = π 5 5 SA = π + πh = π 5 + π The volume is appoximately 5 m, and the suface aea is appoximately 084 m. b) The volumes of the two containes ae equal. c) The fist containe equies less mateial. MHR Pinciples of Mathematics 9 Solutions 67

7 Chapte 9 Section : Investigate Measuement Concepts Chapte 9 Section Question Page 48 a) The question asks you to investigate the dimensions of ectangles that you can fom with a peimete of 4 units. b) Answes will vay. A sample answe is shown. Begin with one gid squae as the width and nine gid squaes as the length. Then, incease the width by one squae and decease the length by the same amount to daw a seies of ectangles with a peimete of 4 units. Chapte 9 Section Question Page 48 a) The question asks you to investigate the dimensions of ectangles that you can fom with a peimete of 0 units. b) Answes will vay. A sample answe is shown. Begin with one toothpick as the width and nine toothpicks as the length. Then, incease the width by one toothpick and decease the length by the same amount to constuct a seies of ectangles with a peimete of 0 units. 68 MHR Pinciples of Mathematics 9 Solutions

8 Chapte 9 Section Question Page 48 a) The question asks you to investigate the dimensions of vaious ectangles with an aea of squae units. b) Answes will vay. A sample answe is shown. Let the space between two pins on the geoboad be unit and use an elastic band to make diffeent ectangles with an aea of squae units. Stat with a width of unit and a length of units. Then, incease the width by unit and decease the length to maintain an aea of squae units. Chapte 9 Section Question 4 Page 48 a) b) The geate the peimete, the highe the cost of the shed, since a geate length of wall is needed. c) Rectangle (a squae) with dimensions 4 m by 4 m will be the most economical. d) Answes will vay. A sample answe is shown. You must conside the type and quality of the mateial used to constuct the shed, and build it with attention to potecting what will be stoed in it. MHR Pinciples of Mathematics 9 Solutions 69

9 Chapte 9 Section Question 5 Page 48 A ectangle with dimensions 4 m by 4 m encloses the geatest aea fo the same amount of fencing. Sketches may vay. A sample sketch is shown. Click hee to load the sketch. Chapte 9 Section Question 6 Page 48 The maximum aea that Colin can enclose is 64 m, using a squae 8 m by 8 m. Click hee to load the speadsheet. 60 MHR Pinciples of Mathematics 9 Solutions

10 Chapte 9 Section Question 7 Page 48 a) Regula polygons enclose the geatest aea. b) Fo a tiangle, the geatest aea is enclosed using an equilateal tiangle with side length m. 44 = 6 + h 08 = h = 6 + h 08 = h 0. 9 h A= bh = 0.9 = 6. 5 The aea of the tiangle is about 6.5 m. Fo a ectangle, the geatest aea is enclosed by a squae with side length 9 m. The aea is 9 9, o 8 m. Fo a hexagon, the geatest aea is enclosed by a egula side length of 6 m. 6 = + h 6 = 9 + h 7 = h 7 = h 50. h A tiangle = bh = = 5. 6 A hexagon = 6A tiangle = = 9. 6 The aea of the hexagon is about 9.6 m. MHR Pinciples of Mathematics 9 Solutions 6

11 Fo a cicle with a cicumfeence of 6 m, the adius is 6, o appoximately 5.7 m. π A = π = π The aea of the cicle is about 0.5 m. c) The shape of the enclosue affects its aea. Diffeent shapes esult in diffeent aeas. The geatest aea can be achieved by using a cicle. 6 MHR Pinciples of Mathematics 9 Solutions

12 Chapte 9 Section Peimete and Aea Relationships of a Rectangle Chapte 9 Section Question Page 487 The maximum aea occus when a squae shape is used. a) 5 m 5 m b) 9 m 9 m c).5 m.5 m d) 0.75 m 0.75 m Chapte 9 Section Question Page 488 a) Answes will vay. Sample answes ae shown. b) The maximum aea occus when a squae shape is used,.5 m by.5 m. Chapte 9 Section Question Page 488 a) The maximum aea occus when a squae shape is used, 0.5 m by 0.5 m. b) The same aea cannot be enclosed using m long baies. It is not possible to ceate a dimension of 0.5 m using m baies. A c) usingope = = A usingbaies = 0 0 = 400 If ope is used, you can enclose , o 0.5 m moe aea. MHR Pinciples of Mathematics 9 Solutions 6

13 Chapte 9 Section Question 4 Page 488 Answes will vay. A speadsheet solution is shown. Let the length epesent the side fomed by the ban. The maximum aea occus with two widths of 4 m and one length of 8 m of fencing. Click hee to load the speadsheet. Chapte 9 Section Question 5 Page 488 a) Use 5 pieces on a side to fom sides that ae.8 5, o 4 m long. A = 4 = 6 9 The maximum aea that can be enclosed is 96 m. b) Use 0 pieces on a side to fom sides that ae.8 0, o 8 m long. A = 8 = 74 8 The maximum aea that can be enclosed is 784 m. 64 MHR Pinciples of Mathematics 9 Solutions

14 Chapte 9 Section Question 6 Page 488 Fom question 4, the maximum aea occus when one length is fomed by the wall, and the length is twice the width. a) Since you need a length that is twice the width, use 0 pieces fo the length, and 5 pieces fo each width, fo dimensions of 8 m by 4 m. A = 4 8 = 9 The existing bode povides 9 96, o 96 m of additional aea. b) Since you need a length that is twice the width, use 0 pieces fo the length, and 0 pieces fo each width, fo dimensions of 56 m by 8 m. A = 8 56 = 568 The existing bode povides , o 784 m of additional aea. MHR Pinciples of Mathematics 9 Solutions 65

15 Chapte 9 Section Question 7 Page 488 Answes will vay. A speadsheet investigation is shown. Click hee to load the speadsheet. The maximum aea of 400 m occus when a squae aea 0 m by 0 m is used. 66 MHR Pinciples of Mathematics 9 Solutions

16 Chapte 9 Section Question 8 Page 489 When 4 sides ae equied, the maximum aea occus when a squae of side length 8 m is used, esulting in an aea of 8, o 64 m. When one side is a hedge, the maximum aea occus when the hedge is used as a length, and the length is twice the width, fo dimensions of 6 m by 8 m, and an aea of 6 8, o 8 m. When a hedge and a fence ae used, the maximum aea occus when a squae is used, fo dimensions of 6 m by 6 m, and an aea of 6, o 56 m. A speadsheet investigation is shown. Click hee to load the speadsheet. MHR Pinciples of Mathematics 9 Solutions 67

17 Chapte 9 Section Question 9 Page 489 a) b) The minimum length of fence occus when the building is used as one length, and the length is twice the width, fo dimensions of m by 6 m. c) The minimum length of fence is 4 m. Chapte 9 Section Question 0 Page 489 Answes will vay. Chapte 9 Section Question Page 489 Answes will vay. Sample answes ae shown. a) A minimum peimete fo a given aea is impotant to know if cost of mateials fo enclosing the aea is a facto, such as fencing in a pastue fo livestock. b) The maximum aea fo a given peimete is impotant to know if space available should be maximized, such as a stoage shed. Chapte 9 Section Question Page 490 Solutions fo the Achievement Checks ae shown in the Teache's Resouce. 68 MHR Pinciples of Mathematics 9 Solutions

18 Chapte 9 Section Question Page 490 The maximum aea occus when a squae is used of side length of appoximately 5.9 m. Investigations may vay. A solution using dynamic geomety softwae is shown. Click hee to load the sketch. Chapte 9 Section Question 4 Page 490 Answes will vay. A speadsheet investigation is shown. The minimum peimete is 0 m using dimensions of 5 m by 0 m. Click hee to load the speadsheet. MHR Pinciples of Mathematics 9 Solutions 69

19 Chapte 9 Section Question 5 Page 490 Answes will vay. An investigation using dynamic geomety softwae is shown. The maximum aea occus when an equilateal tiangle of side length 7. cm is used. Click hee to load the sketch. 60 MHR Pinciples of Mathematics 9 Solutions

20 Chapte 9 Section Question 6 Page 490 Methods will vay. The maximum aea occus when a squae is used of side length of appoximately 4. cm. x + x = x = 400 x = x = 00 x = 00 x 4. 0 x Investigation with dynamic geomety softwae confims the esult. Click hee to load the sketch. MHR Pinciples of Mathematics 9 Solutions 6

21 Chapte 9 Section Question 7 Page 490 Ranjeet is coect. If the sting is used to enclose a cicle, the cicle will have a geate aea than the squae. C = π 4 = π 4 π = π π 8. A = π Chapte 9 Section Question 8 Page 490 Conside the layout of the thee adjoining fields shown. The total length of fence is 6x + 4y = 500. Investigations may vay. A speadsheet investigation is shown. Click hee to load the speadsheet. The maximum aea occus when x = 4.7 m and y = 6.45 m. 6 MHR Pinciples of Mathematics 9 Solutions

22 Chapte 9 Section Minimize the Suface Aea of a Squae-Based Pism Chapte 9 Section Question Page 495 Fom least to geatest suface aea the pisms ae anked B, C, and A. The cubic shape has the least suface aea. The thinnest shape has the geatest suface aea. Chapte 9 Section Question Page 495 a) V = s 5 = s 5 = s 5 = s 8 = s The squae-based pism with the least suface aea is a cube with a side length of 8 cm. b) V = s 000 = s 000 = 000 = s 0 = s s The squae-based pism with the least suface aea is a cube with a side length of 0 cm. c) V = s 750 = s 750 = 750 = s 9. s s The squae-based pism with the least suface aea is a cube with a side length of 9. cm. MHR Pinciples of Mathematics 9 Solutions 6

23 d) V = s 00 = s 00 = s 00 = s 0. 6 s The squae-based pism with the least suface aea is a cube with a side length of 0.6 cm. Chapte 9 Section Question Page 495 a) SA = 6s = 6 8 = 84 The suface aea of the pism is 84 cm. b) SA = 6s = 6 0 = 600 The suface aea of the pism is 600 cm. c) SA = 6s = The suface aea of the pism is about 497 cm. d) SA = 6s = The suface aea of the pism is about 674 cm. Chapte 9 Section Question 4 Page 495 V = s 00 = s 00 = s 4. 7 s The squae-based pism with the least suface aea is a cube with a side length of about 4.7 cm. 64 MHR Pinciples of Mathematics 9 Solutions

24 Chapte 9 Section Question 5 Page 496 a) V = s 4000 = s 4000 = s 5. 9 s The box with the least suface aea is a cube with a side length of about 5.9 cm. b) Answes will vay. Sample answes ae shown. A squae-based pism is difficult to pick up with one hand to pou the laundy soap. Manufactues may also want a lage font on the box to display the company logo and band name. Chapte 9 Section Question 6 Page 496 a) V = s 750 = s 750 = s s The box with the least suface aea is a cube with a side length of 9.09 cm. b) SA = 6s = The minimum aea of cadboad equied is about cm. Chapte 9 Section Question 7 Page L = 500 cm V = s 500 = s 500 = s. 6 s SA = 6s = The minimum aea of cadboad equied is about 0 cm. MHR Pinciples of Mathematics 9 Solutions 65

25 Chapte 9 Section Question 8 Page 496 a) A speadsheet solution is shown. The pism has a base length of 7. cm and a height of 8.5 cm, fo a volume of 500 cm, and a minimum suface aea. Click hee to load the speadsheet. b) The dimensions ae diffeent fom the box in question 7. c) The lidless box equies less mateial. Chapte 9 Section Question 9 Page 496 a) 00mL = 00 cm V = s 00 = s 00 = s 5. 8 s The box with a minimum suface aea is a cube with a side length of 5.8 cm. b) Answes will vay. A sample answe is shown. Cubical boxes ae hade to hold, and the cube would be vey small. c) Answes will vay. Chapte 9 Section Question 0 Page 497 Answes will vay. 66 MHR Pinciples of Mathematics 9 Solutions

26 Chapte 9 Section Question Page 497 You cannot make a cube with an integal side length using all 00 cubes. Find dimensions that ae as close to a cube as possible, such as Chapte 9 Section Question Page 497 a) Pack the boxes as shown. b) This is the closest that 4 boxes can be stacked to fom a cube, which povides the minimum suface aea. c) Answes will vay. A sample answe is shown. Packing 4 boxes pe caton is not the most economical use of cadboad. A cube can be ceated to package 6 tissue boxes: length box ( 4 cm), width boxes ( cm), and height boxes ( 8 cm). Chapte 9 Section Question Page 497 A speadsheet solution is shown. The waehouse should be built with a base length of.6 m and a height of 6. m fo a volume of 000 m and a suface aea of 476. m. Click hee to load the speadsheet. Chapte 9 Section Question 4 Page 497 V = s = s = s 60 = s SA = 6s = 6 60 = 600 The least amount of cadboad equied is 600.0, o 760 cm. MHR Pinciples of Mathematics 9 Solutions 67

27 Chapte 9 Section Question 5 Page 497 V = s 700 = s 700 = s. 9 s SA = 6s = The length of a side is.9 cm. Each flap has a base and a height of. 9, o 4.64 cm. The aea of flaps needed is , o about 4. cm. The total aea of cadboad equied fo a box is , o 05.7 cm. 68 MHR Pinciples of Mathematics 9 Solutions

28 Chapte 9 Section 4 Maximize the Volume of a Squae-Based Pism Chapte 9 Section 4 Question Page 50 The pisms in ode of volume fom geatest to least ae B, C, and A. Chapte 9 Section 4 Question Page 50 a) SA = 6s 50 = 6s 50 6s = = s 5 = s 5 = s The squae-based pism with the maximum volume is a cube with a side length of 5 cm. b) SA = 6s 400 = 6s 400 6s = = s 400 = s 0 = s The squae-based pism with the maximum volume is a cube with a side length of 0 m. c) SA = 6s 750 = 6s 750 6s = = s 5 = s. s MHR Pinciples of Mathematics 9 Solutions 69

29 The squae-based pism with the maximum volume is a cube with a side length of about. cm. 640 MHR Pinciples of Mathematics 9 Solutions

30 d) SA = 6s 00 = 6s 00 6s = = s 00 = s 4. s The squae-based pism with the maximum volume is a cube with a side length of about 4. m. Chapte 9 Section 4 Question Page 50 a) V = s = 5 = 5 The volume is 5 cm. b) V = s = 0 = 8000 The volume is 8000 m. c) V = s =. = 405 The volume is about 405 cm. d) V = s = The volume is about 80 m. MHR Pinciples of Mathematics 9 Solutions 64

31 Chapte 9 Section 4 Question 4 Page 50 A speadsheet solution is shown. The maximum volume occus with a cube of side length 0.8 cm, fo a volume of 60. cm. Click hee to load the speadsheet. Chapte 9 Section 4 Question 5 Page 50 a) SA = 4A + A side bottom ( ) ( = = = 06 ) V = lwh = 6 = 584 The suface aea is 06 cm, and the volume is 584 cm. b) SA = 6s 06 = 6s 06 6s = = s 6 = s 8. s The box with maximum volume is a cube with a side length of 8. cm. c) V = s = The volume of the box in pat b) is about 68 cm, which is geate than the volume of the box in pat a). 64 MHR Pinciples of Mathematics 9 Solutions

32 Chapte 9 Section 4 Question 6 Page 50 a) SA = 4A + A side bottom ( ) ( = = = 67. ) V = lwh = = 5. The suface aea is 6.7 m, and the volume is.5 m. b) SA = 6s 6. 7= 6s 67. 6s = 6 6. = s. = s. s The box with maximum volume is a cube with a side length of. m. c) V = s =. =. The volume of the box in pat b) is. m, which is geate than the volume of the box in pat a). Chapte 9 Section 4 Question 7 Page 50 a) SA = 6s = 6s 6s = 6 6 = s = s 4. s The box with maximum volume is a cube with a side length of.4 m. MHR Pinciples of Mathematics 9 Solutions 64

33 b) V = s = 4. The volume of the box is about m. Chapte 9 Section 4 Question 8 Page 50 a) SA = 6s 500 = 6s 500 6s = = s 50 = s 0. 4 s The box with maximum volume is a cube with a side length of 0.4 cm. b) V = s = The volume of the box is about 8490 cm. Empty Space = V V = = 7790 c) box dive The volume of empty space is 7790 cm. d) Answes will vay. A sample answe is shown. Assume that thee is no empty space in the box. The DVD would fit into the cube with enough oom aound the edges fo the shedded pape. The shedded pape is tightly packed. Chapte 9 Section 4 Question 9 Page 50 Solutions fo the Achievement Checks ae shown in the Teache's Resouce. 644 MHR Pinciples of Mathematics 9 Solutions

34 Chapte 9 Section 4 Question 0 Page 50 a) Dylan has 0 40, o cm of plywood available. SA = 6s = 6s s = = s 4800 = s 69. s Ideally, Dylan needs a cube with a side length 69. cm. b) Diagams will vay. A sample answe is shown. Dylan needs 6 pieces of wood measuing 69. cm by 69. cm. These cannot be cut fom a piece of wood measuing 0 cm by 40 cm. Dylan's closest option is to cut 6 pieces measuing 60 cm by 60 cm, as shown. c) Answes will vay. A sample answe is shown. Assume that Dylan does not want to cut some of the wasted wood, and glue it onto his pieces to make bigge pieces. Assume that the saw cuts ae negligible. MHR Pinciples of Mathematics 9 Solutions 645

35 Chapte 9 Section 4 Question Page 50 a) b) V = s = 0 = 000 The volume of the box is 000 cm. c) d) Assume that the height is half the base length. Fom the diagam, the base length will be 0, o. cm, and the height will be 6.7 cm. V = lwh = The volume of the box is about 85. cm. e) Answes will vay. A sample answe is shown. Assume that the cuts waste a negligible amount of glass. 646 MHR Pinciples of Mathematics 9 Solutions

36 Chapte 9 Section 5 Maximize the Volume of a Cylinde Chapte 9 Section 5 Question Page 508 a) SA = 6π 00 = 6π π 6 π = 6π 00 = π 00 = π 798. h = = The adius of the cylinde is 7.98 cm, and the height is 5.96 cm. b) SA = 6π 0 = 6π 5 0 6π 6 π = 6π 5 = π 5 = π 07. h = 0. 7 = 46. The adius of the cylinde is 0.7 m, and the height is.46 m. MHR Pinciples of Mathematics 9 Solutions 647

37 c) SA = 6π 5= 6π 5 6π = 6π 6π 5 = 6π 5 = 6π 58. h =. 58 = 5. 6 The adius of the cylinde is.58 cm, and the height is 5.6 cm. d) SA = 6π 6400 = 6π π 6 π = 6π 00 = π 00 = π 8. 4 h = 8.4 = The adius of the cylinde is 8.4 mm, and the height is 6.86 mm. 648 MHR Pinciples of Mathematics 9 Solutions

38 Chapte 9 Section 5 Question Page 508 a) V = π h = π The volume of the cylinde is about 9 cm. b) V = π h = π The volume of the cylinde is about m. c) V = π h = π The volume of the cylinde is about 08 cm. d) V = π h = π The volume of the cylinde is about 9 cm. MHR Pinciples of Mathematics 9 Solutions 649

39 Chapte 9 Section 5 Question Page 508 SA = 6π 4 8 = 6π 8 6π 6 π = 6π 4 = π 4 = π 065. h = =. V = π = π h The volume of fuel that the tank can hold is about m. 650 MHR Pinciples of Mathematics 9 Solutions

40 Chapte 9 Section 5 Question 4 Page 508 a) SA = 6π 7 = 6π 6 7 6π π = 6π = π = π 0. h =. 0 = 4. 0 The adius of the cylinde is.0 m, and the height is 4.0 m. b) V = π h = π The volume is about m, o L. c) Answes will vay. A sample answe is shown. Assume that no metal will be wasted in the building pocess, and that no metal is being ovelapped. Chapte 9 Section 5 Question 5 Page 509 a) The height of the optimal cylinde is cm. b) The cylinde will hold, o 60 CDs. 0. c) Answes will vay. A sample answe is shown. Assume that only the dimensions of the CDs need to be consideed, that no exta space is left fo the containe s closing mechanism, and that the plastic containe has negligible thickness. MHR Pinciples of Mathematics 9 Solutions 65

41 Chapte 9 Section 5 Question 6 Page 509 a) Answes will vay. A sample answe is shown. Adjust the suface aea fomula fo the new cylinde, isolate the height and un a few tials using a speadsheet to find the maximum volume. b) SA = π + π h SA π = πh SA π πh = π π SA π π = h The adius and height ae both 7. cm, fo a volume of.9 cm. Click hee to load the speadsheet. 65 MHR Pinciples of Mathematics 9 Solutions

42 Chapte 9 Section 5 Question 7 Page 509 a) Answes will vay. A possible answe is that a cylinde will have the geatest volume. b) SA cylinde = 6π 400 = 6π 400 6π = 6π 6π 400 = π 400 = π. 8 h =. 8 =. 56 SA pism pism = 6s 400 = 6s 400 6s = = s 400 = s 0 = s V = s = 0 = 8000 V cylinde = π h = π The cylinde has a volume of about 908 cm, while the squae-based pism has a volume of 8000 cm. MHR Pinciples of Mathematics 9 Solutions 65

43 Chapte 9 Section 5 Question 8 Page 509 a) Answes will vay. A possible answe is that the sphee will poduce the geatest volume. b) SA = 4π sphee 000 = 4π 000 4π = 4π 4π 500 = π 500 = π. 6 The sphee has a adius of.6 cm. SA cylinde = 6π 000 = 6π 000 6π = 6π 6π 000 = π 000 = π 0. 0 h = 0.0 = The cylinde has a adius of 0.0 cm and a height of 0.60 cm. SAcube = 6s 000 = 6s 000 6s = = s 000 = s 8. 6 s The squae-based pism has a side length of 8.6 cm. 654 MHR Pinciples of Mathematics 9 Solutions

44 c) V sphee 4 = π 4 = π The sphee has a volume of about 849. cm. V V cylinde = s = = π h = π The cylinde has a volume of about cm. cube The squae-based pism has a volume of about cm. d) The sphee has the geatest volume. This will always be the case. e) Fo a given suface aea, volume of a sphee > volume of a cylinde > volume of a squaebased pism. MHR Pinciples of Mathematics 9 Solutions 655

45 Chapte 9 Section 5 Question 9 Page 509 You have m of metal to wok with. a) Fo a cylinde with a top and a bottom, the maximum volume occus fo a adius of 0. cm and a height of 0.6 cm. Click hee to load the speadsheet. b) Fo a cylinde with no top, the maximum volume occus with a adius and height of 0.46 m. Click hee to load the speadsheet. 656 MHR Pinciples of Mathematics 9 Solutions

46 Chapte 9 Section 5 Question 0 Page 509 Methods may vay. A solution using a speadsheet, and anothe using dynamic geomety softwae ae shown. The volume is maximized at 8. cm fo a adius of 6.5 cm and a height of 9.4 cm. + h = 8 + h = 64 4 h = h = h = 56 4 ( ) h = 56 4 Use this fomula fo h, as well as the fomulas fo volume and suface aea, in a speadsheet. Click hee to load the speadsheet. MHR Pinciples of Mathematics 9 Solutions 657

47 Click hee to load the sketch. 658 MHR Pinciples of Mathematics 9 Solutions

48 Chapte 9 Section 6 Minimize the Suface Aea of a Cylinde Chapte 9 Section 6 Question Page 5 a) V = π 00 = π π π = π 600 = π 600 = π 58. h = 5. 8 =. 6 The adius of the cylinde with minimum suface aea is 5.8 cm, and the height is.6 cm. b) V = π = π π = π π = π = π 05. = h = 0.5 = 0. The adius of the cylinde with minimum suface aea is 0.5 m, and the height is.0 m. MHR Pinciples of Mathematics 9 Solutions 659

49 c) V = π 5 = π 5 π = π π 5 = π 5 = π. h =. = 6. 6 The adius of the cylinde with minimum suface aea is. cm, and the height is 6.6 cm. d) V = π 4 = π 4 π π = π = π = π 09. h = 0. 9 = 8. The adius of the cylinde with minimum suface aea is 0.9 m, and the height is.8 m. 660 MHR Pinciples of Mathematics 9 Solutions

50 Chapte 9 Section 6 Question Page 5 a) SA = π + πh = π π The suface aea of the cylinde is about 64 cm. b) SA = π + πh = π π The suface aea of the cylinde is about 5 m. c) SA = π + πh = π. + π The suface aea of the cylinde is about 05 cm. d) SA = π + πh = π π The suface aea of the cylinde is about 5 m. Chapte 9 Section 6 Question Page 5 V = π 540 = π π π = π 70 = π 70 = π 4. 4 h = 4.4 = 8. 8 The adius of the cylinde with minimum suface aea is 4.4 cm, and the height is 8.8 cm. MHR Pinciples of Mathematics 9 Solutions 66

51 Chapte 9 Section 6 Question 4 Page 54 a) V = π 5000 = π π π = π 500 = π 500 = π 9. V = π 000 = π h = 9. = 8. 6 The adius of the cylinde with minimum suface aea is 9. cm, and the height is 8.6 cm. b) Answes will vay. A sample answe is shown. Assume that no exta mateial will be needed to enclose the volume. Chapte 9 Section 6 Question 5 Page = π 6000 = π 6000 = π. 4 π π h =.4 = 4. 8 The adius of the cylinde with minimum suface aea is.4 cm, and the height is 4.8 cm. 66 MHR Pinciples of Mathematics 9 Solutions

52 Chapte 9 Section 6 Question 6 Page 54 a) V = π 75 = π 75 π = π π 75 = π 75 = π 9. h =.9 = 7. 8 The adius of the cylinde with minimum suface aea is.9 cm, and the height is 7.8 cm. b) SA = π + πh = π 9. + π The cost of the aluminum equied is , o $.44. Chapte 9 Section 6 Question 7 Page 54 Answes will vay. A sample answe is shown. It is not always pactical to use cylindes with the optimum volume. They may be hade to use, to handle, to cay, o to stoe. MHR Pinciples of Mathematics 9 Solutions 66

53 Chapte 9 Section 6 Question 8 Page 54 V cylinde = π 500 = π π π = π 50 = π 50 = π 40. V cube 500 = s 500 = s 794. s SA cube = s = 6s = h = 4. 0 = 860. SA cylinde = π + πh = π π A cylinde will have a suface aea of about 49 cm, and a cube will have a suface aea of about 78 cm. A cylinde is moe cost efficient. Chapte 9 Section 6 Question 9 Page 54 The cafeteia does not appea to designed to minimize heat loss The cylindical shape is talle than its diamete. Howeve, thee is a lage glass aea which would encouage sola heating. Chapte 9 Section 6 Question 0 Page 55 Solutions fo the Achievement Checks ae shown in the Teache's Resouce. 664 MHR Pinciples of Mathematics 9 Solutions

54 Chapte 9 Section 6 Question Page 55 a) The minimum suface aea fo the open cylinde occus with a adius of 7.8 cm and a height of 7.8 cm. Click hee to load the speadsheet. b) The minimum suface aea is about 576 cm. c) Answes will vay. A sample answe is shown. Assume that the only cadboad needed is used to enclose the equied volume so thee is no wastage. MHR Pinciples of Mathematics 9 Solutions 665

55 Chapte 9 Section 6 Question Page 55 a) Answes will vay. A possible answe is that the sphee will have the minimum suface aea fo a given volume. b) V cube = s 000 = s 000 = s 0 = s SA cube = 6s = 6 0 = 600 The suface aea of a cube with a volume of 000 cm is 600 cm. V cylinde = π 000 = π 000 π = π π 500 = π 500 = π 54. h = 5. 4 = SA = π + πh cylinde = π π The minimum suface aea of a cylinde with a volume of 000 cm is about 55.7 cm. 666 MHR Pinciples of Mathematics 9 Solutions

56 4 Vsphee = π = π = π 000 = 4π 000 4π = 4π 4π 750 = π 750 = π 6. 0 SA sphee = 4π = 4π The suface aea of a sphee with a volume of 000 cm is about 48. cm. The sphee has the least suface aea. MHR Pinciples of Mathematics 9 Solutions 667

57 Chapte 9 Section 6 Question Page 55 To enclose a maximum volume, use a sphee. SA = 4π 584 = 4π 584 4π = 4π 4π 896 = π 896 = π V 4 = π 4 = π The geatest volume that can be enclosed is about 0 8 cm. 668 MHR Pinciples of Mathematics 9 Solutions

58 Chapte 9 Section 6 Question 4 Page 55 Conside a squae-based pism of base length b and height h inscibed in a cone of adius 0 cm and height 0 cm, as shown. Using simila tiangles, 0 0 h = 0 0.b 5 0 h 5. = 05.b 0 h 05.b 5. = 05.b 05.b 075. b= 0 h h = b Use this elation to investigate the volume of the inscibed squae-based pism. A sample speadsheet is shown. Click hee to load the speadsheet. The maximum volume of 7. cm occus with a base length of 6.67 cm and a height of 0 cm. MHR Pinciples of Mathematics 9 Solutions 669

59 Altenatively, you can use dynamic geomety softwae to investigate the inscibed squae-based pism. A sample sketch is shown, esulting in a simila answe. Click hee to load the sketch. Chapte 9 Section 6 Question 5 Page 55 Use a speadsheet to investigate the suface aea with a constant volume. Solve the volume fomula fo a cone fo h. Calculate the slant height fom the Pythagoean Theoem. The minimum suface aea of 5.4 cm occus with a adius of 4.4 cm and a height of.95 cm. Click hee to load the speadsheet. V = π h V = π h V = π h V π h = π π V = h π s = + h s = + h 670 MHR Pinciples of Mathematics 9 Solutions

60 Chapte 9 Section 6 Question 6 Page 55 Use a speadsheet to investigate the volume with a constant suface aea. Solve the fomula fo the suface aea of a cone to detemine the fomula fo the slant height. Use the Pythagoean theoem to calculate the height of the cone. The maximum volume of cm occus with a adius of 6.9 cm and a height of 9.54 cm. Click hee to load the speadsheet. SA = π + π s SA π = πs SA π πs = π π SA π π = s s = + h s = h h= s MHR Pinciples of Mathematics 9 Solutions 67

61 Chapte 9 Review Chapte 9 Review Question Page 56 a) b) Thee ae 0 possible ectangles, assuming that side lengths ae integes. c) Choose a 0 m by 0 m ectangle in ode to maximize the play aea of the sandbox. Chapte 9 Review Question Page 56 a) b) c) The 4 m by 4 m gaden is the most economical. Fo the same enclosed aea, it has the least peimete. Fewe edging bicks will be equied. 67 MHR Pinciples of Mathematics 9 Solutions

62 Chapte 9 Review Question Page 56 A squae shape has the minimum peimete. Make the whiteboad m by m. Chapte 9 Review Question 4 Page 56 a) The maximum aea occus with a squae of side length 0 cm, fo an aea of 0 0, o 900 m. b) The maximum aea occus with one length equal to twice the width. Use two widths of 0 m each, and one length of 60 m, fo an aea of 0 60, o 800 m. Chapte 9 Review Question 5 Page 56 a) The most economical ink is a squae with a side length of 800, o about 4.4 m. b) Answes will vay. A sample answe is shown. A squae ice ink may not be best as skates may want longe staight uns to gain speed. Chapte 9 Review Question 6 Page 56 A cube measuing about 9.6 cm on a side equies the least amount of mateial. Click hee to load the speadsheet. MHR Pinciples of Mathematics 9 Solutions 67

63 Chapte 9 Review Question 7 Page 56 a) L = 000 cm V = s 000 = s 000 = s 0 = s The box that equies the minimum amount of mateial is a cube with a side length of 0 cm. b) Answes will vay. A sample answe is shown. The suface aea of a cylinde that contains the same volume will be less than the suface aea of the box. The manufactue could save on packaging costs. A cube-shaped box is hade to pick up than a moe ectangula box. Chapte 9 Review Question 8 Page 57 L = 000 cm V = s 000 = s 000 = s 4. 4 s SA = 6s = An aea of about 44 cm of cadboad is equied to make the box. 674 MHR Pinciples of Mathematics 9 Solutions

64 Chapte 9 Review Question 9 Page 57 The maximum volume occus when a cube of side length appoximately 0.58 m is used. Click hee to load the speadsheet. Chapte 9 Review Question 0 Page 57 SA = 6s 00 = 6s 00 6s = = s 00 = s 4. s The maximum volume occus when using a cube with a side length of appoximately 4. cm. Chapte 9 Review Question Page 57 It is not possible to cut six 4. cm by 4. cm pieces fom a 60 cm by 0 cm piece of cadboad. Only fou such pieces fit into these dimensions. MHR Pinciples of Mathematics 9 Solutions 675

65 Chapte 9 Review Question Page 57 The maximum volume of 48.9 cm occus with a adius of 6.8 cm and a height of.6 cm. Click hee to load the speadsheet. Chapte 9 Review Question Page 57 Since thee is no lid, you must change the fomula fo height fom SA π to SA π h= h=. π π Chapte 9 Review Question 4 Page 57 Answes will vay. A sample answe is shown. A cylinde will have a geate volume using the same amount of cadboad, but the squae-based pism may be easie fo customes to stoe. 676 MHR Pinciples of Mathematics 9 Solutions

66 Chapte 9 Review Question 5 Page 57 a) The minimum suface aea is 00.5 cm when the adius is.99 cm, and the height is 8.00 cm. Click hee to load the speadsheet. b) Answes will vay. A sample answe is shown. Assume thee is no waste mateial while making the pop can. Chapte 9 Review Question 6 Page 57 a) The minimum suface aea occus when the height equals the diamete of. cm. The numbe of CDs that the containe will hold is., o b) Answes will vay. A sample answe is shown. Assume that no exta space is allowed inside the containe. c) SA = π + πh = π 6. + π The amount of mateial equied is about 70.4 cm. MHR Pinciples of Mathematics 9 Solutions 677

67 Chapte 9 Chapte Test Chapte 9 Chapte Test Question Page 58 The field should be a squae with a side length of 00 m. Answe B. Chapte 9 Chapte Test Question Page 58 8 L = 8000 cm V = s 8000 = s 8000 = s 0 = s The box should be a cube with side length 0 cm. Answe D. Chapte 9 Chapte Test Question Page 58 The suface aea is a minimum when the diamete equals the height. Answe B. Chapte 9 Chapte Test Question 4 Page 58 SA = 6s 600 = 6s 600 6s = = s 00 = s 0 = s The volume is a maximum when a cube with a side length of 0 cm is used. Answe A. Chapte 9 Chapte Test Question 5 Page 58 The aea is a maximum when a squae shape of side length 50 cm is used. 678 MHR Pinciples of Mathematics 9 Solutions

68 Chapte 9 Chapte Test Question 6 Page 58 Thei volumes of the containes ae equal, since they have the same base aea and the same height. The cylinde equies less mateial to make. Chapte 9 Chapte Test Question 7 Page 58 a) 5 L = 5000 cm V = s 5000 = s 5000 = s 7. s The minimum suface aea occus when a cube of side length appoximately 7. cm is used. b) Answes will vay. A sample answe is shown. Assume that no mateial is ovelapped, and that no exta mateial is equied fo sealing puposes. MHR Pinciples of Mathematics 9 Solutions 679

69 Chapte 9 Chapte Test Question 8 Page 58 a) SA = 6s 8. 64= 6s s = = s 4. 4 = s. = s The maximum volume occus when a cube of side length. m is used. b) V = s =. =. 78 The volume of the box is.78 m. c) The mateial available fo each of the smalle boxes is 864., o.88 m. SA = 6s 88. = 6s 88. 6s = = s 048. = s 069. s Each small box is a cube with a side length of appoximately 0.69 cm. d) V small = s = The total volume of the thee small bins is 0., o 0.99 m. This is less than the volume of the oiginal lage bin. 680 MHR Pinciples of Mathematics 9 Solutions

70 Chapte 9 Chapte Test Question 9 Page 59 The minimum suface aea occus with a adius of 8.5 m and a height of 8.54 m. Click hee to load the speadsheet. Chapte 9 Chapte Test Question 0 Page 59 The maximum volume occus with a base length of. m and a height of 0.5 m. Click hee to load the speadsheet. MHR Pinciples of Mathematics 9 Solutions 68

71 Chaptes 7 to 9 Review Chaptes 7 to 9 Review Question Page 50 a) a = 80 = 68 b + 5 = b = 5 b = 60 b) x + 90 = 80 x = x = 90 x 90 = x = 45 z = = 5 y = = 5 68 MHR Pinciples of Mathematics 9 Solutions

72 Chaptes 7 to 9 Review Question Page 50 a) = = 5 q + 5 = 80 q = 80 5 q = 65 p = 60 p + 70 = 60 p = p = 90 b) b = = 75 c = = 0 d = = 00 Chaptes 7 to 9 Review Question Page 50 a) b) Each exteio angle and its adjacent inteio angle have a sum of 80. Thus an exteio ight angle has an adjacent inteio ight angle. This cannot occu in a tiangle because two ight inteio angles have a sum of 80, leaving no oom fo the tiangle s thid angle. c) d) MHR Pinciples of Mathematics 9 Solutions 68

73 Chaptes 7 to 9 Review Question 4 Page 50 a) ( n ) 80 = 44n 80n 60 = 44n 80n n= 44n n 6n = 60 6n 60 = 6 6 n = 0 The polygon has 0 sides. b) The sum of the exteio angles is 60º fo all polygons. Chaptes 7 to 9 Review Question 5 Page 50 a) b) Answes will vay. A sample answe is shown. You can use dynamic geomety softwae to otate a line segment about one of its endpoints five times though an angle of 60º. Then, join the endpoints of the line segments fomed. 684 MHR Pinciples of Mathematics 9 Solutions

74 Chaptes 7 to 9 Review Question 6 Page 50 Adam is coect. The median fom the hypotenuse divides the aea of a ight tiangle into two equal pats. You can veify this conjectue using dynamic geomety softwae. A sample sketch is shown. Click hee to load the sketch. Chaptes 7 to 9 Review Question 7 Page 50 a) It is false that the diagonals of a paallelogam ae equal in length. A counte-example is shown. b) It is tue that the line segment joining the midpoints of two sides of a tiangle is always paallel to the thid side. You can use dynamic geomety softwae to show that inteio angles add to 80º, making the line segments paallel. c) It is false that the diagonals of a tapezoid ae neve equal in length. A counte-example is shown. MHR Pinciples of Mathematics 9 Solutions 685

75 Chaptes 7 to 9 Review Question 8 Page 50 a) c c = = c =. c =. c 58. P = =. 9 A = = 8. The peimete is.9 m, and the aea is 8. m. b) 5 = a = a = a 0 = a 0 = a a 7. P = = 60. A = 8 7. = The peimete is 60. cm, and the aea is 55.7 cm. 686 MHR Pinciples of Mathematics 9 Solutions

76 Chaptes 7 to 9 Review Question 9 Page 50 P = = 0. 0 A= A A ectangle cutout = = = The peimete is 0.0 m, and the aea is 8.56 m. Chaptes 7 to 9 Review Question 0 Page 5 a) c = c = c =. 0 c =. 0 c 6. SA = A + A + A + A base left side bottom ight side = = V = A h base = = 5. 6 The suface aea is appoximately 48. m, and the volume is 5.6 m. MHR Pinciples of Mathematics 9 Solutions 687

77 b) SA = A + 4A base tiangle = = = 45 V = Abase h = The suface aea is 45 cm, and the volume is appoximately 458. cm. Chaptes 7 to 9 Review Question Page 5 5 ml = 5 cm 5 = π. 6 h V = π h 5 =. 96π h 5. 96π h =. 96π. 96π 5 = h. 96π 80. h The height of the can is 8.0 cm. 688 MHR Pinciples of Mathematics 9 Solutions

78 Chaptes 7 to 9 Review Question Page 5 a) s s s = 8 + = = 7 s = 7 s 85. SA = πs + π = π π 08 The aea of pape equied is about 08 cm. b) V = π h = π 8 75 The volume of the cone is appoximately 75 cm. Chaptes 7 to 9 Review Question Page 5 a) V 4 = π 4 = π 0 50 The volume of the golf ball is appoximately 50 mm. b) SA = 4π = 4π The suface aea of the golf ball is appoximately 507 mm. c) The entie suface of a golf ball is coveed with small indentations (commonly known as dimples). Due to the pesence of dimples, the actual suface aea of the golf ball is geate and the volume of the golf ball is less than that calculated in pats a) and b). MHR Pinciples of Mathematics 9 Solutions 689

79 Chaptes 7 to 9 Review Question 4 Page 5 a) Allie should make a squae gaden, using pieces, o 6.5 m, on a side. b) The aea of the gaden is 65., o 4.5 m c) The peimete of the gaden is , o 6 m. Chaptes 7 to 9 Review Question 5 Page 5. V = s = s = s. 5 s SA = 6s = The aea of cadboad equied is about 774 cm. Chaptes 7 to 9 Review Question 6 Page 5 a) SA = 6s 50 = 6s 50 6s = = s 5 = s 5 = s The maximum volume occus with a cube of side length 5 cm. b) The maximum volume of 4 cm occus with a adius of.8 cm and a height of 5.6 cm. Click hee to load the speadsheet. 690 MHR Pinciples of Mathematics 9 Solutions

80 Chaptes 7 to 9 Review Question 7 Page 5 The minimum suface aea of about 9 cm occus with a adius of.9 cm and a height of 7.9 cm. Click hee to load the speadsheet. MHR Pinciples of Mathematics 9 Solutions 69