Apply differentiation to practical problems — Paper 2 · IGCSE Mathematics - Additional 0606

6 The volume of a closed cylinder of base radius x cm and height h cm is 500 cm3. (i) Express h in terms of x. [1] 2 1000 2 (ii) Show that the total surface area of the cylinder is given by A = 2r x + cm . [2] x (iii) Given that x can vary, find the stationary value of A and show that this value is a minimum. [5]

8 marks

Mark scheme: 6(i) 2 500 B1 Ignore units Condone r for x πx h = 500 → h = πx 2 6(ii) A = 2 πx 2 + 2 πxh M1 Correct expression for A and insert for their h. 2 500 2 1000 A1 Answer given = 2 πx + 2 πx × = 2 πx + 2 Condone r for x . πx x 6(iii) Differentiate: at least one power reduced by 1 M1 d A 1000 A1 = 4πx − d x x 2 dA 1000 A1 = 0 → x = 3 isw or ( x = 4.3 ( 0 ) ) dx 4π 2 1000 2 A1 awrt 349 A = 2π( 4.3 ) + = 349 cm 4.3 d 2 A 2000 B1 Correct second differential (need = 4 π + ( > 0 ) or a positive value not be evaluated) and conclusion. 2 3 dx x or ( → min) Examine correct gradient either side of x = 4.3 and conclusion

This question in 0606/22 Oct/Nov 2017

11 h r A container is a circular cylinder, open at one end, with a base radius of r cm and a height of h cm. The volume of the container is 1000 cm3. Given that r and h can vary and that the total outer surface area of the container has a minimum value, find this value. [8]

8 marks

Mark scheme: 11 1000 1000 B1 h = 2 or r = soi πr πh 2  1000  M1 S = πr + 2πr  2  oe or  πr   1000 1000 ( h ) oe S = π  + 2π  π h  πh 2  1000  A1 S = πr + 2   or better or  r  1000 1000 2 ) S = + 2π ( h 1 h π dS −2 B2 B1 FT for each term correct = 2πr − 2000 r or dr − d S −2 1 = − 1000 h + 1000π h 2 d h dS 3 1000 M1 = 0, r = oe or dr π d S 32 1000 = 0, h = oe d h π 2 M1  1000  2000 S = π  3  + or  π  1000 3 π 1 1000  1000  2 S = + 2 1000π  3  1000  π  3 π 439 or 439.3 to 439.4 A1

This question in 0606/22 Feb/March 2020

12 (a) Find the x-coordinates of the stationary points of the curve y = e 3 x ( 2x + 3) 6 . [6] (b) A curve has equation y = f( x) and has exactly two stationary points. Given that f ll ( x) = 4 x - 7 , f l ( 0.5) = 0 and f l ( 3) = 0, use the second derivative test to determine the nature of each of the stationary points of this curve. [2] (c) In this question all lengths are in centimetres. h x 4x The diagram shows a solid cuboid with height h and a rectangular base measuring 4x by x. The volume of the cuboid is 40 cm3. Given that x and h can vary and that the surface area of the cuboid has a minimum value, find this value. [5]

13 marks

Mark scheme: 12(a) 3 x B1 d e ( ) 3 x = 3e d x 6 M1 d ( 2 x + 3) 5 = k ( 2 x + 3) dx their (3e3x) (2x + 3)6 + M1 (e3x) (their 12 (2x + 3)5) (3e3x) (2x + 3)6 + (e3x) (12 (2x + 3)5) A1 (3e3x) (2x + 3)5 (2x + 7) = 0 M1 x = –1.5, –3.5 A1 12(b) x = 0.5 f ′′ ( 0.5 )[ = −5 ] < 0  max B2 B1 for either one correct x = 3 f ′′ ( 3 )[ = 5 ] > 0  min 12(c) 10 B1 h = x 2 2  10  B1 S = 8 x + 10 x  their 2   x  d S − 2 M1 = 16 x − 100 x oe d x 2 25 A1 d S 16 x − 100 x− = 0, x = 3 oe FT their = 0 if possible 4 d x 81.4 or 81.4325… rot to four or more A1 figs

This question in 0606/21 May/June 2020

9 x m B C D 300 m A E 400 m The rectangle ABCDE represents a ploughed field where AB = 300 m and AE = 400 m . Joseph needs to walk from A to D in the least possible time. He can walk at 0.9 ms -1 on the ploughed field and at 1.5 ms -1 on any part of the path BCD along the edge of the field. He walks from A to C and then from C to D. The distance BC = x m . (a) Find, in terms of x, the total time, T s, Joseph takes for the journey. [3] (b) Given that x can vary, find the value of x for which T is a minimum and hence find the minimum value of T. [6]

9 marks

Mark scheme: 9(a) 2 2 B1 ( AC = ) 300 + x seen isw 2 2 M1 using clearly indicated value for their 300 + x time for AC = oe AC or their CD 0.9 400 − x or time for CD = oe 1.5 2 2 A1 300 + x 400 − x T = + oe seen 0.9 1.5 isw 9(b) − 1 B2 accept unsimplified; 2 2 if incorrect allow B1 for correct dT 1 ( 300 2 + x 2 ) = × 2 x − oe 1 dx 2 0.9 3 2 2 ± 2 differentiation of 300 + x ( ) dT M1 d T set their = 0 must be a function of x dx d x 2 2 2 A1 equation in x2 with square root removed 25 x = 9 300 + x oe ( ) x = 225 (m) A1 T = 533 (s) or 1600/3 (exact value) A1 or 8 min 53 s

This question in 0606/23 Oct/Nov 2020

11 In this question all lengths are in centimetres. 4 3 2 The volume and surface area of a sphere of radius r are rr and 4rr respectively. 3 x y The diagram shows a solid object made from a hemisphere of radius x and a cylinder of radius x and height y. The volume of the object is 500 cm3. (a) Find an expression for y in terms of x and show that the surface area, S, of the object is given by 5 2 1000 S = r x + . [4] 3 x (b) Given that x can vary and that S has a minimum value, find the value of x for which S is a minimum. [4]

8 marks

Mark scheme: 11(a) 4 3 2 M1 500 = πx + πx y oe 6 1  4 3  A1 if first M0, SC1 for oe, isw 500 − y =  πx  1  4 3  6   πx 2 y = 2  500 − πx  oe seen πx  3  2 2  500 2  M1 dep on first M1 S = 2πx + πx + 2πx  2 − x   πx 3  Correct completion to given answer: A1 5 2 1000 S = πx + 3 x 11(b) 10 1000 B2 B1 for each term Differentiates S: πx − oe 3 x 2 10 1000 M1 d S πx − = 0 and attempt to solve FT their providing at least B1 awarded 2 3 x d x 300 A1 x = 3 isw or 4.57[07...] nfww π

This question in 0606/22 May/June 2021

11 The volume, V, of a cone with base radius r and vertical height h is given by rr 2 h . 3 The curved surface area of a cone with base radius r and slant height l is given by rrl . A cone has base radius r cm, vertical height h cm and volume V cm3. The curved surface area of the cone is 4r cm2. 2 16 2 (a) Show that h = 2 - r . [4] r (b) Show that V = r 16r 2 - r 6 . [2] 3 (c) Given that r can vary and that V has a maximum value, find the value of r that gives the maximum volume. [5]

11 marks

Mark scheme: 11(a) 4 B1 l = r h 2 = l 2 − r 2 or l 2 = r 2 + h 2 M1 2 2 M1 FT their l ; dep on previous M1 2 4 2 4 2 2 = h =  − r or  r + h r r 2 16 2 2 2 or l = and h = l − r r 2 2 16 2 A1 Correct, convincing completion to h = − r r 2 Alternative method 2 2 (B1) l = r + h 2 2 (M1) πr r + h = 4π 2 (M1) 2 4 2 2 ( r + h ) = r 2 16 2 (A1) Correct, convincing completion to h = − r r 2 11(b) π 2 16 2 M1 r 2 − r 3 r A1 π 4  16 2  r  2 − r  3  r  π 2 6 and correct completion to 16r − r 3 111(c) − d V π  1 2 6  5 B3 B2 for 2 1 =  (16 r 32 r − 6 r ) oe − r )  ( 2 6 − d r 3  2 16 r − r 32 r − 6 r   k ( 2 5 ) ( ) where k is a constant and k ≠ 0 or B1 for 1 2 6 − 2 k 16 r − r × ( f ( r ) ) where ( ) f(r) ≠ 32 r − 6 r 5 d V M1 FT their f(r) = ar +br5 Equates their to 0 and solves as far as for a, b ≠ 0 d r r4 = … r = 1.52 or 1.519[67…] rot to 4 or more sf or A1 2 oe 4 3

This question in 0606/21 Oct/Nov 2021

12 In this question all lengths are in centimetres. O R 12 P Q C h A B 8 The diagram shows a right triangular prism of height h inside a right pyramid. The pyramid has a height of 12 and a base that is an equilateral triangle, ABC, of side 8. The base of the prism sits on the base of the pyramid. Points P, Q and R lie on the edges OA, OB and OC, respectively, of the pyramid OABC. Pyramids OABC and OPQR are similar. 3 3 2 (a) Show that the volume, V, of the triangular prism is given by V = ( ah + bh + ch ) where a, 9 b and c are integers to be found. [4] (b) It is given that, as h varies, V has a maximum value. Find the value of h that gives this maximum value of V. [3]

7 marks

Mark scheme: 12(a) PQ 12 − h M1 = oe 8 12 2  or  12 − h = Area Δ PQR oe  12  16 3 8(12 − h ) A1 PQ = oe 12  12 − h  2 or Area ΔPQR = 16 3    12 oe 1  8(12 − h )  2 π M1 FT their PQ or Area ΔPQR providing of correct V = ×   × sin × h oe structure 2  12  3 2  or 16 3  12 − h × h  12  3 3 2 A1 V = ( h − 24 h + 144 h ) 9 12(b) d V 3 2 B1 3 3 2 = ( 3h − 48 h + 144 ) FT their V = ( h − 24 h + 144 h ) if of the d h 9 9 same structure 3 2 M1 their ( 3h − 48 h + 144 ) = 0 and must be a 3-term quadratic; must be an attempt 9 at a derivative factorises/solves 4 oe identified as the only solution; cao A1

This question in 0606/22 Feb/March 2022

12 In this question all lengths are in centimetres. x y A container is a half‑cylinder, open as shown. It has length y and uniform cross‑section of radius x. The volume of the container is 25 000. Given that x and y can vary and that the outer surface area, S, of the container has a minimum value, find this value. [8]

8 marks

Mark scheme: 12 πx 2 y M1  25000 oe 2 50000 50000 A1 y  or πxy  πx 2 x  50000  2 M1 k S  πx  2   πx FT their y providing of the form 2  πx  π x 50000 2 A1 S   πx x d S  2 M1 k 2  50000 x  2π x oe FT from expression of form  mx d x x Equates to zero and solves for x: A1 25000 x  3 or 19.9647 … π 2 DM1 dep on previous M1; 50000  25000  S min   π  3  FT their positive value of x used in a 25000  π  correct expression for S 3 π 3760 or 3756 to 3757 A1

This question in 0606/23 May/June 2022

8 In this question all lengths are in centimetres. The volume of a cylinder with radius r and height h is rr 2 h and its curved surface area is 2rrh . 4 3 2 The volume of a sphere with radius r is rr and its surface area is 4rr . 3 h r The diagram shows a solid object in the shape of a cylinder of base radius r and height h, with a hemisphere of radius r on top. The total surface area of the object is 300 cm 2. (a) Find an expression for h in terms of r. [2] 5 3 (b) Show that the volume, V, of the object is 150r - r r . [3] 6

10 marks

Mark scheme: 8(a) 2πrh + 2πr 2 + πr 2  = 300  oe M1 300 − 3πr 2 A1 h = oe, isw 2πr 8(b) 2  300 − 3πr 2  2 3 M2 FT their h providing in terms of r and V = πr  their  + πr oe derived from a dimensionally correct  2πr  3 equation in (a); M1 for 2  300 − 3πr 2  3 V = πr  their  + kπr ,  2πr  2 k ≠ oe 3 Correct completion to given answer: A1 5 3 150 r − πr nfww 6 8(c) 5 2 B1 Derivative of V : 150 − πr oe, soi, isw 2  5 2  M1 dV their  150 − πr  = 0 and solves as far FT their providing that at least one  2  d r term is correct as r = … 300 A1 r = oe or 4.37 (cm) 5π 5 3 M1 FT their 4.37 150(their 4.37) − π(their 4.37) 6 437 or awrt 437 (cm3) isw A1

This question in 0606/22 Oct/Nov 2022

5 The curved surface area of a cylinder with radius r and height h is 2rrh . A closed cylinder has radius r cm and volume 1000 cm 3. 2 2000 2 (a) Show that the total surface area of the cylinder is 2rr + cm . [3] r (b) Find the value of r which makes this area a minimum. You should show that your value of r gives a minimum for this area. [5]

8 marks

Mark scheme: 5(a) Correct use of r 2 h = 1000 to find an expression that M2 M1 for r 2 h = 1000 soi can be used to eliminate h 1000 1000 e.g. h = or πrh = πr 2 r Correct substitution and completion to given answer A1 2 1000 2 2000 e.g. 2πr + 2πr  = 2πr + πr 2 r 5(b) 2000 B2 B1 for one correct term Correct derivative: 4πr − oe, isw r 2 2000 M1 FT their derivative providing that at 4πr − = 0 and solves for r r 2 least one term is correct 2000 A1 r = 3 oe, isw 4π or 5.42 or 5.419[26…] rot to 3 or more dp d 2 S –3 A1 Dep on previous mark Second derivative = 4π + 4000r and d r 2 d 2 S When r = 5.42, > 0 oe [hence minimum] d r 2 or 4π + 4000(5.42) –3 > 0 oe [hence minimum] d 2 S or = 12 or 37 to 38 [hence minimum] d r 2 d 2 S or as r  0 , > 0 [hence minimum] d r 2 OR correctly finds the values of the first derivative at 5.42 oe  h, where h is small [hence minimum]

This question in 0606/23 Oct/Nov 2023

11 In this question all lengths are in centimetres. A B x D C The diagram shows a rectangle ABCD with BC = x . The area of the rectangle is 400 cm2. x Two identical quarter-circles of radius 2, with centres A and C, are removed from the rectangle to make the shaded shape. Given that x can vary, find the value of x that gives the minimum value of the perimeter of the shaded shape and hence find this minimum value. [7]

7 marks

Mark scheme: 11  πx   400 x  B2 400 P = 2    x  2    oe B1 for rectangle length = soi  4   x 2  x Correct first derivative B1 FT their P providing it is of the form π 800 a   bx oe with a an integer and b a 2 x 2 x constant d P M1 FT provided one term of their Equates their to 0 and solves for x derivative is correct d x 40 A1 x = or 22.6 π or 22.56[75...] π  40 800 M1 FT their value of x provided it is P =    greater than 0 2  π  40 π P = 70.9 or 70.89[81...] A1

This question in 0606/21 May/June 2024