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Math16A - Fall 01 Final Solutions

Math16A - Fall 01 Final Solutions - SOLUTIONS TO MATH 16A...

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SOLUTIONS TO MATH 16A, AUTUMN 2001 FINAL 1. Use logarithmic differentiation to compute f 0 ( x ) /f ( x ) = d dx (ln( f ( x ))) = d dx (ln(( x + 1) 1 - x )) = d dx ((1 - x ) ln( x + 1)) = - ln( x + 1) + 1 - x x + 1 Thus, f 0 (2) = f (2)( - ln(2 + 1) + 1 - 2 2+1 ) = 3 - 1 ( - ln(3) + ( - 1 / 3)) = - ln(3) 3 - 1 9 . 2. a. A, b. G, c. E, d. B, e. C, corrected 3. Write f ( t ) = # of bacteria in tens of thousands t hours after noon . We know that f is an exponential function. So, we may write f ( t ) = Ae rt for some constants A and r . We know 3 = f (0) = A and 4 = f (1) = 3 e r so that e r = 4 / 3. Hence, g (3) = 3 e r 3 = 3(4 / 3) 3 = 3(64 / 27) = (64 / 9). That is, there are (64 / 9) × 10 4 = 71 , 111 1 9 individuals at 3pm. bf 4. Z 2 1 (2 x 3 - 5 e - 7 x + 4 x - 8 x ) dx = [ 1 2 x 4 + 5 7 e - 7 x + 4 5 x 5 4 - 8 ln( x )] | 2 1 = (8 + 5 7 e - 14 + 4 5 2 5 4 - 8 ln(2)) - ( 1 2 + 5 7 e - 7 + 4 5 - 8 ln(1)) = 8 + 5 7 e - 14 + 8 5 4 2 - 8 ln(2) - 1 2 - 5 7 e - 7 - 4 5 = 6 . 7 + 5 7 e - 14 + 8 5 4 2 - 8 ln(2) - 5 7 e - 7 5. a. J, b. B, c. D, d. J, e. F 6. We write the price in dollars. The demand function is linear, so we have x ( p ) = mp + b for some parameters m and b . We compute m = 500 - 350 . 8 - 1 = - 750 and 350 = - 750(1) + b so that b = 1100. The revenue is then R ( p ) = px ( p ) = - 750 p 2 + 1100 p . Differentiating, R 0 ( p ) = - 1500 p + 1100 so that we maximize revenue with R 0 ( p ) = 0 or p = 11 / 15.

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