Mathematics Grade 12 P1 NSC Supplementary 2012 Memo (Eng)

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 12 MATHEMATICS P1 FEBRUARY/MARCH 2012 MEMORANDUM MARKS: 150 This memorandum consists of 20 pages. Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 2 DBE/Feb.–Mar. 2012 NSC – Memorandum QUESTION 1 1.1.1 3x 2 − 5 x = 2 3x 2 − 5x − 2 = 0 9 standard form (3 x + 1)( x − 2) = 0 9 factors 1 9 both answers x = − or x = 2 (3) 3 1.1.2 2 x− =5 x x − 2 = 5x 2 x2 − 5x − 2 = 0 9 standard form − b ± b 2 − 4ac x= 2a − (−5) ± 25 − 4(1)(−2) 9 subs in correct x= formula 2(1) 5 ± 33 x= 2 99 answers x = 5,37 or x = −0,37 (one for each answer) (4) 1.1.3 ( x + 1)( x − 3) > 12 9 multiplication x 2 − 2 x − 3 > 12 x 2 − 2 x − 15 > 0 ( x − 5)( x + 3) > 0 9 factors + 0 − 0 + OR –3 5 –3 5 x < −3 or x > 5 99 answer (4) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 3 DBE/Feb.–Mar. 2012 NSC – Memorandum 1.2 r+ p=2 r = 2− p 9 r = 2− p 6r + 5rp − 5 p = 8 9 substitution 6(2 − p) + 5(2 − p) p − 5 p = 8 12 − 6 p + 10 p − 5 p 2 − 5 p = 8 9 simplification 5 p2 + p − 4 = 0 (5 p − 4)( p + 1) = 0 4 9 factors p= or p = −1 9 p-answers 5 ⎛4⎞ r = 2 − ⎜ ⎟ or r = 2 − (−1) ⎝5⎠ 6 99 r-answers r= r =3 (7) 5 OR r+ p=2 9 p = 2−r p = 2−r 6r + 5rp − 5 p = 8 6r + 5r (2 − r ) − 5(2 − r ) = 8 9 substitution 6r + 10r − 5r 2 − 10 + 5r = 8 5r 2 − 21r + 18 = 0 (5r − 6)(r − 3) = 0 9 standard form 9 factors 6 r= or r = 3 9 r-answers 5 ⎛6⎞ p = 2 − ⎜ ⎟ or p = 2 − (3) ⎝5⎠ 99 p-answers 4 p= p = −1 (7) 5 1.3 Let the shortest side be x 9 let the shortest side be x Sides of the prism: x ; 2x ; 3x Volume = lbh 9 x ; 2x ; 3x (x)(2x)(3x) = 3 072 6 x 3 = 3 072 9 (x)(2x)(3x) = 3 072 x 3 = 512 9 answer x = 3 512 x=8 (4) [22] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 4 DBE/Feb.–Mar. 2012 NSC – Memorandum QUESTION 2 2.1 Tn = a + (n − 1)d 9d=4 173 = −7 + (n − 1)(4) 9 Tn = −7 + 4(n − 1) 173 = −7 + 4n − 4 4n = 184 n = 46 9 answer (3) OR Tn = 4n − 11 99 Tn = 4n − 11 173 = 4n − 11 9 answer 4n = 184 (3) n = 46 2.2 n S n = [a + l ] 2 9 subs of n = 46 46 9 subs of a and l = [−7 + 173] into the correct 2 formula = 23 [166 ] = 3 818 9 answer (3) OR n S n = [2a + (n − 1)d ] 2 46 9 subs of n = 46 = [2(−7) + (45)(4)] 9 subs of a and d 2 into the correct = 23 [−14 + 180 ] formula = 3 818 9 answer (3) 2.3 46 9n=1 ∑ (4n − 11) n =1 9 top value = 46 9 4n – 11 (3) [9] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 5 DBE/Feb.–Mar. 2012 NSC – Memorandum QUESTION 3 3.1.1 1 1 r=− 9 r=− 2 2 ⎛ 1⎞ 9 answer T4 = 1⎜ − ⎟ (2) ⎝ 2⎠ 1 =− 2 3.1.2 ⎛ 1⎞ n −1 ⎛ 1⎞ n Tn = 4⎜ − ⎟ Tn = −8⎜ − ⎟ n −1 ⎝ 2⎠ ⎝ 2⎠ ⎛ 1⎞ 9 4⎜ − ⎟ n −1 ⎛ 1⎞ n ⎝ 2⎠ 1 ⎛ 1⎞ 1 = 4⎜ − ⎟ = −8⎜ − ⎟ 9 substitution 64 ⎝ 2⎠ 64 ⎝ 2⎠ n −1 n 1 ⎛ 1⎞ 1 ⎛ 1⎞ = ⎜− ⎟ OR = ⎜− ⎟ n −1 256 ⎝ 2 ⎠ 256 ⎝ 2 ⎠ 1 ⎛ 1⎞ 9 = ⎜− ⎟ ⎛ 1⎞ 8 ⎛ 1⎞ n −1 8 ⎛ 1⎞ ⎛ 1⎞ n −1 256 ⎝ 2 ⎠ ⎜− ⎟ = ⎜− ⎟ ⎜− ⎟ = ⎜− ⎟ ⎝ 2⎠ ⎝ 2⎠ ⎝ 2⎠ ⎝ 2⎠ 8 = n −1 8 = n −1 n=9 n=9 9 answer (4) OR 1 T4 = − 2 1 T5 = 9 T5 and T6 4 1 9 T7 T6 = − 8 1 T7 = 9 T8 16 1 T8 = − 32 1 9 answer T9 = 64 (4) n=9 3.1.3 a S∞ = 9 substitution into 1− r 4 correct formula = ⎛ 1⎞ 9 answer 1− ⎜ − ⎟ ⎝ 2⎠ (2) 8 = 3 Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 6 DBE/Feb.–Mar. 2012 NSC – Memorandum 3.2 For a geometric sequence: x +1 x − 3 T2 T3 = 9 = 1 x +1 T1 T2 x + 2x + 1 = x − 3 2 x2 + 2x + 1 = x − 3 x2 + x + 4 = 0 9 standard form x2 + x + 4 = 0 2 − 1 ± 1 − 4(1)(4) ⎛ 1 ⎞ 15 x= OR ⎜x + ⎟ + =0 2(1) ⎝ 2⎠ 4 9 subs in quadratic 2 formula − 1 ± − 15 ⎛ 1 ⎞ 15 15 x= ⎜x + ⎟ + ≥ >0 2 ⎝ 2⎠ 4 4 9 non-real/no x- Solution is non-real. values There is no x-value that makes the sequence geometric. (4) OR T2 T3 9 = For a geometric sequence: T1 T2 x +1 x − 3 = 1 x +1 9 standard form x + 2x + 1 = x − 3 2 x2 + x + 4 = 0 9 subs in discriminant b 2 − 4ac = 1 − 4(1)(4) = −15 9 non-real/no x- Solution is non-real values There is no x-value that makes the sequence geometric. (4) [12] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 7 DBE/Feb.–Mar. 2012 NSC – Memorandum QUESTION 4 4.1 17 10 5 2 9 complete pattern –7 –5 –3 2 2 9r=1 r=1 9s=2 s=2 (3) OR NOTE: Candidates may do 4.2 first. d (n) = n 2 − 10n + 26 9 r = d (5) d (n) = n 2 − 10n + 26 = (5) − 10(5) + 26 2 =1 9r=1 s = d (6) = (6) − 10(6) + 26 9s=2 2 (3) = 12 OR d (n) = (n − 5) + 1 2 r = d (5) = (5 − 5) + 1 2 9 d (n) = (n − 5) + 1 2 =1 s = d (6) 9r=1 = (6 − 5) + 1 2 9s=2 =2 (3) 4.2 2a = 2 a =1 9a=1 3a + b = −7 9 method ∴ 3(1) + b = −7 b = −10 ∴ a + b + c = 17 9 b = – 10 1 − 10 + c = 17 9 c = 26 c = 26 (4) ∴ d (n) = n 2 − 10n + 26 OR Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 8 DBE/Feb.–Mar. 2012 NSC – Memorandum a + b + c = 17 4a + 2b + c = 10 a + b + c = 17 9 method 3a + b = −7 4a + 2b + c = 10 9a= 1 9a + 3b = −21 9a + 3b + c = 5 9 c = 26 9a + 3b + c = 5 3a + b = −7 9 b = – 10 (4) − 21 + c = 5 5a + b = −5 c = 26 2a = 2 OR a + b = −9 a =1 4a + 2b = −16 3(1) + b = −7 2a + 2b = −18 b = −10 2a = 2 (1) − 10 + c = 17 a =1 c = 26 b = −10 d (n) = n 2 − 10n + 26 d (n) = n 2 − 10n + 26 OR 9 method 2a = 2 9a= 1 a =1 c = 26 9 c = 26 d (n) = n + bn + 26 2 17 = (1) 2 + b + 26 b = −10 d (n) = n 2 − 10n + 26 9 b = – 10 (4) OR n −1 d(n) = [2(first difference) + (n − 2)(second difference)] + d(1) 2 n −1 d ( n) = [2(−7) + (n − 2)(2)] + 17 9 method 2 9a=1 n −1 9 c = 26 d ( n) = [− 18 − 2n] + 17 9 b = – 10 2 (4) d (n) = (n − 1)(−9 − n) + 17 d (n) = n 2 − 10n + 26 OR Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 9 DBE/Feb.–Mar. 2012 NSC – Memorandum (n − 1)(n − 2) d (n) = (n − 1)d (2) − (n − 2)d (1) + second difference × 2 2(n − 1)(n − 2) 9 method d (n) = (n − 1)(10) − (n − 2)(17) + 9a=1 2 9 c = 26 d (n) = 10n − 10 − 17n + 34 + (n − 1)(n − 2) 9 b = – 10 d (n) = n 2 − 10n + 26 (4) OR 9 method d (n) = (n − 5) + 1 2 9a=1 = n 2 − 10n + 26 9 c = 26 a =1 9 b = – 10 b = −10 (4) c = 26 4.3 d (8) = (8) 2 − 10(8) + 26 9 subs t = 8 OR By symmetry = 10 m d (8) 9 answer = d (5 + 3) (2) = d (5 − 3) 9 method = d ( 2) = 10 9 answer (2) OR 17, 10, 5, 2, 1, 2, 5, 10 9 method so d(8) = 10 9 answer (2) 4.4 Since the distance from P is decreasing for n < 5 the athlete is moving towards P. 99 decreasing Since the distance from P is increasing for n > 5, the athlete is moving away from Moving towards P. 99 increasing Moving away OR It is sufficient to show that d is decreasing when n < 5 and increasing when n > 5 d (n) = n 2 − 10n + 26 9 d ′(n) = 2n − 10 d ′(n) = 2n − 10 d ′(n) = 2(n − 5) For n < 5, 2(n − 5) < 0 9 2(n – 5) < 0 d ′(n) < 0 ∴ decreasing 9 decreasing For n > 5, 2(n − 5) > 0 d ′(n) > 0 ∴ increasing 9 increasing (4) [13] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 10 DBE/Feb.–Mar. 2012 NSC – Memorandum y D QUESTION 5 C f g B G x O H g A 5.1 x=0 9 answer y=0 9 answer (2) 5.2 f ( x) = −2 x + 8 x + 10 2 9 equate to 0 x 2 − 4x − 5 = 0 ( x − 5)( x + 1) = 0 9 factors x = 5 or x = – 1 9 x-values H(5 ; 0) 9 answer (4) 5.3 f ( x) = −2 x + 8 x + 10 2 9 method f ( x) = −2( x − 2) 2 + 18 9 ( x − 2) 2 Range of f is y ≤ 18 OR y ∈ (−∞ ; 18] 9 18 9 answer OR (4) f ( x) = −2 x 2 + 8 x + 10 8 x=− 2(−2) 9 method x=2 9 x=2 y = −2(2) 2 + 8(2) + 10 9 y = 18 y = 18 9 answer Range of f is y ≤ 18 OR y ∈ (−∞ ; 18] (4) OR 9 method Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 11 DBE/Feb.–Mar. 2012 NSC – Memorandum 5 −1 9 x=2 x= 2 x=2 9 y = 18 y = −2(2) 2 + 8(2) + 10 9 answer y = 18 (4) Range of f is y ≤ 18 OR y ∈ (−∞ ; 18] OR f ( x) = −2 x 2 + 8 x + 10 f ′( x) = −4 x + 8 9 method 0 = −4 x + 8 9 x=2 x=2 9 y = 18 y = −2(2) 2 + 8(2) + 10 9 answer y = 18 (4) Range of f is y ≤ 18 OR y ∈ (−∞ ; 18] 5.4 f (1) = −2(1) 2 + 8(1) + 10 9 substitution f(1) f (1) = 16 9 substitution g(1) 16 (2) g (1) = 1 g (1) = 16 C(1 ; 16) lies on both the graphs of f and g OR 16 9 equating − 2 x 2 + 8 x + 10 = x 9 answer − 2 x + 8 x + 10 x − 16 = 0 3 2 (2) x − 4x − 5x + 8 = 0 3 2 ( x − 1)( x 2 − 3x − 8) = 0 x = 1 or x 2 − 3x − 8 = 0 C(1 ; 16) 5.5 p ( x) = f (3 x) 9 3x = 2 3x = 2 NOTE: 2 9x= Answer Only: Full Marks 3 2 x= 9 y = 18 3 (3) 2 TP ( ; 18) 3 OR 24 9 x=− p ( x) = −2(3x) 2 + 8(3x) + 10 2(−18) = −18 x 2 + 24 x + 10 2 9x= 3 Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 12 DBE/Feb.–Mar. 2012 NSC – Memorandum 24 9 y = 18 x=− (3) 2(−18) 2 x= 3 2 TP ( ; 18) 3 OR p ( x) = −2(3 x) 2 + 8(3 x) + 10 = −18 x 2 + 24 x + 10 9 0 = −36 x + 24 p ′( x) = −36 x + 24 0 = −36 x + 24 2 9x= 2 3 x= 9 y = 18 3 (3) 2 TP ( ; 18) 3 9 OR 2 ⎛ 2⎞ p( x) = −18⎜ x − ⎟ + 18 p( x) = −2(3x) 2 + 8(3 x) + 10 ⎝ 3⎠ = −18 x 2 + 24 x + 10 2 2 9x= ⎛ 2⎞ 3 = −18⎜ x − ⎟ + 18 ⎝ 3⎠ 9 y = 18 2 (3) TP ( ; 18) [15] 3 Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 13 DBE/Feb.–Mar. 2012 NSC – Memorandum QUESTION 6 6.1 f ( x) = 3 x 9 answer f −1 ( x) = log 3 x (1) 6.2 y −1 f ( x) = log 3 x (Log Graph) 9shape f 9x-intercept f ( x) = 3 x (Exponential Graph) 9shape (0 ; 1) f -1 9y-intercept (4) x (1 ; 0) 6.3 x>0 99 answer (2) OR x ∈ (0 ; ∞ ) 6.4 0<x ≤ 1 9 critical values 9 notation (2) 6.5 y > −4 OR y ∈ (−4 ; ∞) 99 answer (2) 6.6 g ( x) = −3 x−2 OR g ( x) = − f ( x − 2) 9 – (sign) 9 x–2 OR (2) g ( x) = − 1 x 9 ( ) 1 3 OR g ( x) = − f ( x) 9 9 – (sign) 1 9 9 (2) [13] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 14 DBE/Feb.–Mar. 2012 NSC – Memorandum QUESTION 7 7.1 88 9 × 850 000 88 100 × 850 000 = R 748 000 100 9 answer OR (2) 12 × 850 000 9 100 12 = R102 000 × 850 000 100 Loan amount = R850 000 – R102 000 9 answer = R748 000 (2) 7.2 ⎛ 0,09 ⎞ 12 1 + ie = ⎜1 + ⎟ 0,09 ⎝ 12 ⎠ 9 12 ie = 0,09380689 9 substitution r = 9,38% p.a 9 answer ≠ 9,6% 9 decision Not correct (4) OR i 9 ⎛ i ⎞ 12 12 1 + 0,096 = ⎜1 + ⎟ 9 substitution ⎝ 12 ⎠ 12 1,096 = 1 + i 12 i 1,007668183 = 1 + 12 i = 0,092018201 9 answer r = 9,2% p.a 9 decision ≠ 9% (4) Not correct 7.3 x[1 − (1 + i ) − n ] Pv = 9 subs into i correct ⎡ ⎛ 0,09 ⎞ − 240 ⎤ formula x ⎢1 − ⎜1 + ⎟ ⎥ ⎢⎣ ⎝ 12 ⎠ ⎥⎦ 0,09 748 000= 9i= 0,09 12 9 n = – 240 12 9 answer x = R 6 729,95 (4) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 15 DBE/Feb.–Mar. 2012 NSC – Memorandum 7.4 x[1 − (1 + i ) − n ] Pv = i ⎡ ⎛ 0,09 ⎞ −n ⎤ 9 subs into 7 000⎢1 − ⎜1 + ⎟ ⎥ ⎢⎣ ⎝ 12 ⎠ ⎥⎦ correct 748 000= formula 0,09 12 −n 561 ⎛ 0,09 ⎞ = 1 − ⎜1 + ⎟ 700 ⎝ 12 ⎠ −n ⎛ 0,09 ⎞ 139 9 simplification ⎜1 + ⎟ = ⎝ 12 ⎠ 700 ⎛ 0,09 ⎞ 139 − nlog⎜1 + ⎟ = log 9 use of logs ⎝ 12 ⎠ 700 n = 216,35 months = 18,03 years 9 answer (4) OR x[1 − (1 + i ) −12n ] Pv = i ⎡ ⎛ 0,09 ⎞ −12n ⎤ 9 subs into 7 000 ⎢1 − ⎜1 + ⎟ ⎥ correct ⎢⎣ ⎝ 12 ⎠ ⎥⎦ 748 000= formula 0,09 12 −12 n 561 ⎛ 0,09 ⎞ = 1 − ⎜1 + ⎟ 700 ⎝ 12 ⎠ −12 n 9 simplification ⎛ 0,09 ⎞ 139 ⎜1 + ⎟ = ⎝ 12 ⎠ 700 ⎛ 0,09 ⎞ 139 9 use of logs − 12nlog⎜1 + ⎟ = log ⎝ 12 ⎠ 700 n = 18,03 years 9 answer (4) [14] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 16 DBE/Feb.–Mar. 2012 NSC – Memorandum QUESTION 8 8.1 f ( x) = 9 − x 2 f ( x + h) = 9 − ( x + h) 2 9 substitution = 9 − x 2 − 2 xh − h 2 f ( x + h) − f ( x) = −2 xh − h 2 9 simplification − 2 xh − h 2 f ′( x) = lim h→0 h 9 formula h(−2 x − h) = lim 9 common factor h→0 h = lim(−2 x − h) h→0 = −2 x 9 answer (5) OR f (x + h ) − f (x ) 9 formula f ′( x ) = lim h→0 h = lim 9 − (x + h ) − 9 − x 2 2 ( ) 9 substitution h→0 h = lim ( ) 9 − x + 2 xh + h 2 − 9 + x 2 2 9 simplification h→0 h − 2 xh − h 2 = lim h→0 h h(− 2 x − h ) 9 common factor = lim h→0 h = lim(− 2 x − h ) h→0 9 answer = −2 x (5) 8.2.1 Dx [1 + 6 x ] ⎡ 1 ⎤ 1 = Dx ⎢1 + 6 x 2 ⎥ 9 6x 2 ⎢⎣ ⎥⎦ 9 answer 1 − (2) = 3x 2 8.2.2 8 − 3x6 9 x −5 y= 8 x5 3 9 x 1 3 8 = 5− x x 8 3 9 − 5 x −6 = x −5 − x 3 8 9 − dy 3 8 = −5 x − 6 − (4) dx 8 [11] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 17 DBE/Feb.–Mar. 2012 NSC – Memorandum QUESTION 9 y g A(− 1 ; 0) C(3 ; 0) x O D(0 ; − 6) f E B 9.1 f ( x) = a ( x + 1) 2 ( x − 3) 99 substitution of x-values − 6 = a (0 + 1) 2 (0 − 3) 9 subs (0 ; – 6) − 6 = −3a a=2 9a=2 f ( x) = 2( x 2 + 2 x + 1)( x − 3) 9 simplification = 2 x 3 − 2 x 2 − 10 x − 6 (5) 9.2 f ′( x ) = 6 x 2 − 4 x − 10 9 f ′( x) = 6 x 2 − 4 x − 6 9 f ′( x) = 0 6 x 2 − 4 x − 10 = 0 3x 2 − 2 x − 5 = 0 9 factors (3x − 5)( x + 1) = 0 5 9 x-value x= or x = −1 3 9 y-value ⎛5 512 ⎞ (5) B⎜ ; − ⎟ OR B(1,67 ; − 18,96) ⎝3 27 ⎠ 9.3 h( x) = 2 x 3 − 2 x 2 − 10 x − 6 − (6 x − 6) = 2 x 3 − 2 x 2 − 16 x 9 h( x) = 2 x 3 − 2 x 2 − 16 x 9 h′( x) = 6 x 2 − 4 x − 16 h′( x) = 6 x 2 − 4 x − 16 9 h′( x) = 0 0 = 3x 2 − 2 x − 8 0 = (3x + 4)( x − 2) 9 factors 4 x=− or x = 2 3 4 9 correct x-value ∴x=− 3 (5) [15] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 18 DBE/Feb.–Mar. 2012 NSC – Memorandum QUESTION 10 10.1 y = 5(1) − 8 9 subs 1 = −3 (1) Point of contact is (1 ; – 3) 10.2 − 3 = 2(1) 3 + p(1) 2 + q (1) − 7 9 subs (1 ; – 3) 2= p+q g ′( x) = 6 x 2 + 2 px + q 9 g ′(1) = 5 g ′( x) = 6 x 2 + 2 px + q 5 = 6(1) 2 + 2 p(1) + q 9 subs x = 1 and y = 5 −1 = 2 p + q 9 simplification 9 p-value p = −3 9 q-value q=5 (6) [7] QUESTION 11 y 9 x-intercepts 9 turning point 99 shape [4] f 1 /2 2 3 x −1 − /3 1 Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 19 DBE/Feb.–Mar. 2012 NSC – Memorandum QUESTION 12 12.1 x, y ∈ N 0 x + y ≤ 100 y ≤ − x + 100 99 x + y ≤ 100 x ≤ 50 OR x ≤ 50 99 x + 2 y ≤ 180 9 x ≤ 50 x + 2 y ≤ 180 1 y ≤ − x + 90 (5) 2 12.2 999 each constraint Large tables 170 160 150 9 feasible region 140 (4) 130 120 110 100 90 A(20 ; 80) 80 70 60 50 FEASIBLE REGION 40 30 20 10 Small Tables 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210 220 230 12.3 90 tables 9 answer (1) 12.4 P = 300 x + 400 y 9 answer (1) 12.5 Maximum at A (20 ; 80) 99 answer 20 small tables and 80 large tables. (2) 12.6 P = qx + 400 y q m=− q 400 9 m=− q 1 400 −1 ≤ − ≤− 9 20 0 ≤ q ≤ 400 400 2 20 0 ≤ q ≤ 400 (2) [15] TOTAL: 150 Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 20 DBE/Feb.–Mar. 2012 NSC – Memorandum QUESTION 12.2 Large tables 170 160 150 140 130 120 110 100 90 A(20 ; 80) 80 70 60 50 40 FEASIBLE REGION 30 20 10 Small tables 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210 220 230 Copyright reserved