Ch 21 Practice Problems [48 marks]

Transcription

1 Ch 21 Practice Problems [4 marks] A dog food manufacturer has to cut production costs. She ishes to use as little aluminium as possible in the construction of cylindrical cans. In the folloing diagram, h represents the height of the can in cm and x, the radius of the base of the can in cm. The volume of the dog food cans is 600 cm. 1a. Sho that h = 600. πx = πx 2 h 600 πx 2 = h (AG) (A1) Note: Aard for correct substituted formula, (A1) for correct substitution. If anser given not shon aard at most (A0). Find an expression for the curved surface area of the can, in terms of x. Simplify your anser. 1b. C = 2πx 600 πx 2 C = 1200 x 1 (or 1200x ) (A1) Note: Aard for correct substitution in formula, (A1) for correct simplification. [??? marks] Hence rite don an expression for A, the total surface area of the can, in terms of x. 1c.

2 A = 2π x x 1 (A1)(A1)(ft) Note: Aard (A1) for multiplying the area of the base by to, (A1) for adding on their anser to part (b) (i). For both marks to be aarded anser must be in terms of x. [??? marks] Differentiate A in terms of x. 1d. [ marks] da dx = 4πx 1200 x 2 (A1)(ft)(A1)(ft)(A1)(ft) Notes: Aard (A1) for 4πx, (A1) for 1200, (A1) for x 2. Aard at most (A2) if any extra term is ritten. Follo through from their part (b) (ii). [??? marks] Find the value of x that makes A a minimum. 1e. [ marks] πx = 0 x 2 x 1200 = 4π x = 4.57 (or equivalent) Note: Aard for using their derivative, for setting the derivative to zero, (A1)(ft) for anser. Follo through from their derivative. Last mark is lost if value of x is zero or negative. [ marks] 1f. Calculate the minimum total surface area of the dog food can. A = 2π(4.57 ) (4.57) 1 A = 94 Note: Follo through from their ansers to parts (b) (ii) and (d).

3 A shipping container is to be made ith six rectangular faces, as shon in the diagram. The dimensions of the container are length 2x idth x height y. All of the measurements are in metres. The total length of all telve edges is 4 metres. Sho that y =12 x. 2a. [ marks] 4(2x) + 4y + 4x = 4 Note: Aard for setting up the equation. 12x + 4y = 4 Note: Aard for simplifying (can be implied). y = 4 12x 4 y = 12 x (AG) (A1) Note: The last line must be seen for the (A1) to be aarded. [ marks] x + y = 12 Sho that the volume V m of the container is given by 2b. 2 V = 24x 6x V = 2x x (12 x) (A1) Note: Aard for substitution into volume equation, (A1) for correct substitution. = 24x 2 6x (AG) Note: The last line must be seen for the (A1) to be aarded. 2c. Find. dv dx

4 dv dx = 4x 1 (A1)(A1) Note: Aard (A1) for each correct term. x 2 Find the value of x for hich V is a maximum. 2d. [ marks] 4x 1 x 2 = 0 Note: Aard for using their derivative, for equating their anser to part (c) to 0. for sketch of V = 24x 2 6x, for the maximum point indicated dv dx for sketch of = 4x 1x 2, for the positive root indicated (,, ) Note: Follo through from their part (c). [ marks] Find the maximum volume of the container. 2e. V = 24 ( 2 ) 6 ( ) Note: Aard for substitution of their value from part (d) into volume equation. 56.9( m 512 )(, 56...) 9 Note: Follo through from their anser to part (d). 2f. Find the length and height of the container for hich the volume is a maximum. [ marks] length = 16 (A1)(ft)(G1) Note: Follo through from their anser to part (d). Accept 5.4 from use of 2.67 height = 12 ( ) = 4 Notes: Aard for substitution of their anser to part (d), (A1)(ft) for anser. Accept.99 from use of [ marks] The shipping container is to be painted. One litre of paint covers an area of 15 m 2. Paint comes in tins containing four litres. 2g. Calculate the number of tins required to paint the shipping container. [4 marks]

5 SA = SA = 4 ( ) Note: Aard for substitution of their values from parts (d) and (f) into formula for surface area (m ) ( (m )) (A1) Note: Accept 92.5 ( ) from use of sf ansers Number of tins = (= 1.54) 15 4 [4 marks] Note: Aard for division of their surface area by tins required (A1)(ft) Note: Follo through from their ansers to parts (d) and (f). A parcel is in the shape of a rectangular prism, as shon in the diagram. It has a length l cm, idth cm and height of 20 cm. The total volume of the parcel is 000 cm. a. Express the volume of the parcel in terms of l and. [1 mark] 20l V = 20l (A1) [1 mark] b. Sho that l = = 20l Note: Aard for equating their anser to part (a) to 000. l = Note: Aard for rearranging equation to make l subject of the formula. The above equation must be seen to aard. 150 = l Note: Aard for division by 20 on both sides. The above equation must be seen to aard. l = 150 (AG)

6 c. The parcel is tied up using a length of string that fits exactly around the parcel, as shon in the folloing diagram. Sho that the length of string, S cm, required to tie up the parcel can be ritten as 00 S = , 0 < 20. S = 2l (20) Note: Aard for setting up a correct expression for S ( ) (20) Notes: Aard for correct substitution into the expression for S. The above expression must be seen to aard. = (AG) d. The parcel is tied up using a length of string that fits exactly around the parcel, as shon in the folloing diagram. Dra the graph of S for 0 < 20 and 0 < S 500, clearly shoing the local minimum point. Use a scale of 2 cm to represent 5 units on the horizontal axis (cm), and a scale of 2 cm to represent 100 units on the vertical axis S (cm).

7 (A1)(A1)(A1)(A1) Note: Aard (A1) for correct scales, indo and labels on axes, (A1) for approximately correct shape, (A1) for minimum point in approximately correct position, (A1) for asymptotic behaviour at = 0. Axes must be dran ith a ruler and labeled and S. For a smooth curve (ith approximately correct shape) there should be one continuous thin line, no part of hich is straight and no (one-to-many) mappings of. The S-axis must be an asymptote. The curve must not touch the S-axis nor must the curve approach the asymptote then deviate aay later. [4 marks] e. The parcel is tied up using a length of string that fits exactly around the parcel, as shon in the folloing diagram. [ marks] Find. ds d

8 d (A1)(A1)(A1) 1 Notes: Aard (A1) for 4, (A1) for 00, (A1) for or 2. If extra terms present, aard at most (A1)(A1)(A0). 2 [ marks] f. The parcel is tied up using a length of string that fits exactly around the parcel, as shon in the folloing diagram. Find the value of for hich S is a minimum ds 4 = 0 = 4 = d Note: Aard for equating their derivative to zero. =.66 ( 75, ) Note: Follo through from their anser to part (e). g. The parcel is tied up using a length of string that fits exactly around the parcel, as shon in the folloing diagram. [1 mark] Write don the value, l, of the parcel for hich the length of string is a minimum. 17.(, ) (A1)(ft) Note: Follo through from their anser to part (f). [1 mark]

9 h. The parcel is tied up using a length of string that fits exactly around the parcel, as shon in the folloing diagram. Find the minimum length of string required to tie up the parcel Note: Aard for substitution of their anser to part (f) into the expression for S. = 110 (cm) ( , ) Note: Do not accept 109. Follo through from their ansers to parts (f) and (g). International Baccalaureate Organization 2015 International Baccalaureate - Baccalauréat International - Bachillerato Internacional Printed for Bolton High School