Unformatted text preview: x . Setting P = 0 yields x = 10 . Observe that P 00 = − 12 < for all x , so this gives a maximum pro f t. Hence P max = (60 + 20) (150 − 30) = $9600 . 21. Observe f rst that 90 mi / h = 90 60 · 88 = 132 ft / s . Let a ( t ) be the acceleration at any time t after braking, v ( t ) be the corresponding velocity, and d ( t ) the distance travelled while braking. Deceleration implies a ( t ) = − 44 , so v ( t ) = − 44 t + C 1 . Substitute t = 0 to get v (0) = C 1 ⇒ C 1 = 132 . So v ( t ) = − 44 t + 132 , which gives d ( t ) = − 22 t 2 + 132 t + C 2 . Since d (0) = 0 , we have d (0) = C 2 ⇒ C 2 = 0 . So d ( t ) = − 22 t 2 + 132 . The car comes to a stop when v ( t ) = 0 ⇒ 44 t = 132 ⇒ t = 3 s . Hence the distance covered is s (3) = − 22 (3) 2 + 132 (3) = 198 ft ....
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 Fall '09
 RAdin
 Harshad number, $2, 90 mi, $9600, 132 ft, 198 ft

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