HW 12 Solutions - x . Setting P = 0 yields x = 10 . Observe...

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Mathematics 408 K: Homework 12 Gagan Tara Nanda 08 th November, 2004 HW 12 : # s 3 , 5 , 21 3. Since the company produces a total of x + y cars and wants to f ll an order for 150 cars, we have x + y = 150 y = 150 x . Substitute this into the expression for C to get C = x 2 + 2 (150 x ) 2 +2 x (150 x )+190 x + 30 (150 x )+50 . There is no need to expand this out. We wish to minimize C ,sod i f erentiate (use the Product Rule where appropriate) to get: C 0 =2 x + 4 (150 x )( 1) + 2 (150 x )+2 x ( 1) + 190 + 30 ( 1) =2 x 140 . Setting C 0 =0 ,weget x =70 .S in c e C 00 =2 > 0 for all x , this leads to a minimum cost. 5. The key idea in this problem (and # s 6 , 7 ) is the following: Let the room rate increase by x multiples of $2 . Then the number of rooms occupied decreases by 3 x . So the pro f t function is given by the cost per room times the number of rooms occupied minus the cost of cleaning the occupied rooms, which is P =( 6 6+ 2 x ) (150 3 x ) 6(150 3 x ) =( 6 0+ 2 x ) (150 3 x ) . Then P 0 = 2 (150
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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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