Calc Final Solutions - 3 2 3 2 1 lim(3 x − 5 x 3 x 11 = 3...

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Unformatted text preview: 3 2 3 2 1. lim(3 x − 5 x + 3 x + 11) = 3 ⋅ 2 − 5 ⋅ 2 + 3 ⋅ 2 + 11 = 21 x →2 2. x 2 − 10 x + 24 x 2 − 6 x − 4 x + 24 = lim 2 x →9 x 2 − 13 x + 36 x →9 x − 4 x − 9 x + 36 x( x − 6) − 4( x − 6) ( x − 6)( x − 4) = lim = lim > x →9 x ( x − 4) − 9( x − 6) x →9 ( x − 4)( x − 9) x−6 = lim x →9 x − 9 x−6 lim = −∞ x →9− x − 9 x−6 lim =∞ x →9+ x − 9 x 2 − 10 x + 24 Thus, lim 2 does not exist. x →9 x − 13 x + 36 lim 3. Sent to you 4. R = 20x – 0.01x2 R ′ = 20 – 0.01(2)x = 20 – 0.02x R ′ (1000) = 20 – 0.02(1000) = 20 – 20 = 0 The marginal profit at x = 1000 is 0. 5. Sent to you 6. C = 4 x3 7 x4 + 3 (4 x3 )′(7 x 4 + 3) − 4 x3 (7 x 4 + 3)′ (7 x 4 + 3) 2 C′ = 12 x 2 (7 x 4 + 3) − 4 x 3 ⋅ 28 x 3 84 x 6 + 36 x 2 − 112 x 6 = = (7 x 4 + 3) 2 (7 x 4 + 3) 2 −28 x 6 + 36 x 2 4 x 2 (7 x 4 − 9) = =− (7 x 4 + 3) 2 (7 x 4 + 3) 2 7. f = 70(6 x 6 − x + 9)14 f ′ = 70 ⋅14(6 x 6 − x + 9)13 ⋅ (6 x 6 − x + 9)′ = 980(6 x 6 − x + 9)13 (36 x5 − 1) 8. P = 16 x − 0.2 x 2 − 300 P′ = 16 − 0.2(2) x = 16 − 0.4 x P′ = 0 ⇒ 16 − 0.4 x = 0 x = 40 9. C = 500 + 6 x + 0.08 x 2 The average cost is __ 500 + 6 x + 0.08 x 2 500 C ( x) = = + 6 + 0.08 x x x 500 C ′ = (−1)500 x −2 + 0.08 = − 2 + 0.08 x 500 C′ =0 ⇒ − 2 + 0.08 = 0 x 2 0.08 x = 500 x 2 = 6250 x = 79.06 C (79.06) = 10. C = 500 + 300 x 2 + x 3 R = 9000 x − 90 x 2 The profit is P = R − C = 9000 x − 90 x 2 − (500 + 300 x 2 + x 3 ) = 9000 x − 90 x 2 − 500 − 300 x 2 − x 3 = − x 3 − 390 x 2 + 9000 x − 500 P′ = −3x 2 − 780 x + 9000 P′ = 0 ⇒ −3 x 2 − 780 x + 9000 = 0 __ ( x must be positive) 500 + 6 + 0.08(79.06) = 18.649 79.06 x 2 + 260 x − 3000 = 0 −260 ± 2602 − 4(1)(−3000) −260 ± 79600 = 2 2 −260 ± 282.13472 = 2 x = −271.067 or x = 11.0674 x= Check: P(11) = $49979 P(12) = $49612 11. PV = ∫ 20000e 0 5 −0.08t dt = 20000 ∫ e −0.08t dt 0 5 e −0.08 t 5 = 20000 = 20000 −12.5e 0 −0.08 0 −0.08 t 5 = −250000 e −0.08t = −250000(e −0.4 − e0 ) 0 5 = −250000(0.67032 − 1) = 82419.99 12. FV = ∫ 30000e0.07(10 − t ) dt = 30000 ∫ e 0.7 − 0.07 t dt 0 0 10 10 = 30000 ∫ e0.7 e −0.07 t dt = 30000e0.7 ∫ e −0.07 t dt 0 0 10 10 e −0.07 t −0.07 t 10 = 60412.58121 = −863036.8745 e 0 −0.07 0 = −863036.8745(e −0.7 − e0 ) = −863036.8745(0.4965853 − 1) = −863036.8745(0.4965853 − 1) = 434465.45 13. y1 = − x 2 − 3x + 4 y2 = x 2 − 3 x − 4 10 Find the x-coordinate of the points of intersection − x 2 − 3x + 4 = x 2 − 3x − 4 8 = 2 x2 4 = x2 x = ±2 The area between two curves is A = ∫ (− x 2 − 3 x + 4 − ( x 2 − 3 x − 4))dx −1 1 = ∫ (− x 2 − 3 x + 4 − x 2 + 3 x + 4)dx −1 1 1 2 = ∫ (−2 x 2 + 8)dx = − x 3 + 8 x 3 −1 −1 2 2 = − ⋅ 13 + 8 ⋅ 1 − − ⋅ (−1)3 + 8 ⋅ (−1) 3 3 2 2 2 2 = − + 8 − − 8 = − + 8 − + 8 3 3 3 3 44 = 3 1 14. C ′ = 20 x + 100 20 dx x + 100 u = x + 100 du = dx C=∫ 20 du = 20ln | u | + K u = 20ln | x + 100 | + K Find K: C (0) = 20ln | 0 + 100 | + K = 0 20ln |100 | + K = 0 K = −20ln(100) Then C = 20ln | x + 100 | −20ln(100) C (60) = 20ln | 60 + 100 | −20ln(100) = 20(ln160 − ln100) 160 = 20ln = 20ln1.6 100 = 9.4 =∫ 15. f = 6000 − 7000e −0.01x − 2000e −0.03 y ∂f = −7000(−0.01)e −0.01x = 70e −0.01x ∂x ∂f ∂x = 70e −0.01(130) = 19.077 x =130 ...
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This note was uploaded on 03/16/2011 for the course MATH 1600 taught by Professor Calc during the Spring '11 term at Macomb Community College.

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