HW11sol

# HW11sol - HW11 Solutions Math 1A Spring 2009 with Professor...

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HW11 Solutions Math 1A Spring 2009 with Professor Stankova § 4.7 #52, 54, 56, 58, 62 52. We want to maximize the derivative of y = 1 + 40 x 3 - 3 x 5 . So, to maximize y 0 = 120 x 2 - 15 x 4 , ﬁrst take the derivative of y 0 , and ﬁnd the critical numbers. y 00 = 240 x - 60 x 3 = 60 x (4 - x 2 ) = 60 x (2 - x )(2 + x ) = 0 So, critical numbers are x = 0 , - 2 , 2. We do not have a closed interval here, so if we check to see where y 0 is increasing and decreasing (by looking at the sign of y 00 = ( y 0 ) 0 ), we see that the only possible places for y 0 to be a maximum are at x = - 2 , 2. interval ( -∞ , - 2) ( - 2 , 0) (0 , 2) (2 , ) sign of ( y 0 ) 0 + - + - Now y 0 (2) = y 0 ( - 2) = 240, so we get y 0 at a maximum for both x -values. To ﬁnd the ac- tual points on the curve, y (2) = 1+40(2) 3 - 3(2) 5 = 225, and y ( - 2) = 1+40( - 2) 3 - 3( - 2) 5 = - 223, so the points are (2 , 225) and ( - 2 , - 223). 54. (a) If P ( x ) = R ( x ) - C ( x ) is at a maximum, then P 0 ( x ) = R 0 ( x ) - C 0 ( x ) = 0, so R 0 ( x ) = C 0 ( x ) (b) P ( x ) = xp ( x ) - C ( x ) = 1700 x - 7 x 2 - 16000 + 500 x - 1 . 6 x 2 + 0 . 004 x 3 . To maximize this, ﬁrst ﬁnd P 0 ( x ) and determine critical numbers. P 0 ( x ) = 1700 - 14 x - 500+3 . 2 x - 0 . 012 x 2 = - 0 . 012 x 2 - 10 . 8 x +1200 To ﬁnd where P 0 ( x ) = 0, use the quadritic formula, and we see the critical numbers are x = 100 , - 1000. We only take values of x 0 since x is the amount produced. So make a table to check the sign of P 0 ( x ): interval (0 , 100) (100 , ) sign of P 0 ( x ) + - From this, we can conclude that a production level of x = 100 does give the maximum proﬁt. 56. (a) For \$10 each, Terry sells 20 necklaces per day and for \$11 each, he can sell 18. So p (20) = 10 and p (18) = 11. If p = mx + b , then m = 10 - 11 20 - 18 = - 1 2 . Plug in values to ﬁnd b : 10 = - 1 2 20 + b = - 10 + b

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So b = 10 + 10 = 20 and p ( x ) = - 1 2 x + 20. (b) C ( x ) = 6 x P ( x ) = xp ( x ) - C ( x ) = - 1 2 x 2 + 20 x - 6 x P 0 ( x ) = - x + 20 - 6 = - x + 14 = 0 gives x = 14 as the only critical number. Check the sign of P 0 ( x ) now and we can see that x = 14 maximizes P ( x ) (proﬁt). interval (0 , 14) (14 , ) sign of P 0 ( x ) + - Now the problem actually asked for the selling price to maximize proﬁt. x = 14 just tells us that he needs to sell that many per day. So ﬁnd the selling price: p (14) = - 1 2 14 + 20 = - 7 + 20 = \$13 58. Need to maximize R ( x ) = xp ( x ). First ﬁnd p ( x ). p (99) = 810 and p (100) = 800. Let p ( x ) = mx + b m = 810 - 880 99 - 100 = - 10 1 = - 10 Plugging in one of the points then to solve for b gives p ( x ) = - 10 x + 1800. So now we need to maximize R ( x ) = - 10 x 2 + 1800 x . R 0 ( x ) = - 20 x + 1800 = 0 gives x = 90 as the only critical number. Checking the sign of R 0 ( x ) shows that x = 90 maximizes R ( x ). interval
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## This note was uploaded on 03/30/2010 for the course MATH 11111 taught by Professor Stankova during the Fall '09 term at Berkeley.

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HW11sol - HW11 Solutions Math 1A Spring 2009 with Professor...

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