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Unformatted text preview: wtm369 Homework 11 Helleloid (58250) 1 This printout should have 14 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points What is the minimum value of the product P = xy when y = 3 + 4 x ? 1. minimum value = 3 4 2. minimum value = 9 16 correct 3. minimum value = 13 16 4. minimum value = 9 16 5. minimum value = 3 4 6. minimum value = 13 16 Explanation: When y = 3 + 4 x , P = xy = x (3 + 4 x ) = 3 x + 4 x 2 . Now the minimum value of P will occur at a critical point of P , i.e. , at the solution, x , of P ( x ) = 3 + 8 x = 0 , in other words when x = 3 8 . Since P ( x ) = 8 > , we thus see that minimum value = 9 16 . 002 10.0 points A Pine Car Derby car rolls along a track with velocity v ( t ) = 3 + 2 t t 2 feet per second. If it starts at time t = 0, after how many seconds will its position be maximized? 1. t = 3 seconds correct 2. t = 0 seconds 3. t = 9 seconds 4. t = 2 seconds 5. none of these 6. t = 1 second Explanation: Since we are maximizing the position, x ( t ), of the car, this will occur when x ( t ) = 0, i.e. , when x ( t ) = v ( t ) = 3 + 2 t t 2 = 0 . But 3 + 2 t t 2 = (3 t )(1 + t ) = 0 when t = 3 and when t = 1. But t = 1 is not allowed because the car starts at time t = 0. Now x ( t ) = v ( t ) = 2 2 t , so x (3) = 0. Hence the position of the car will be maximized after t = 3 seconds. 003 10.0 points A rectangular dog pound with three kennels as shown in the figure wtm369 Homework 11 Helleloid (58250) 2 consists of a rectangular fenced area divided by two partitions. Determine the maximum possible area of this pound if 48 yards of chain link fencing is available for its construction. 1. max area = 74 sq.yards 2. max area = 72 sq.yards correct 3. max area = 76 sq.yards 4. max area = 73 sq.yards 5. max area = 75 sq.yards Explanation: Let the dimensions of the floor of the dog pound be as shown in the figure below x y Then the area of the pound is given by A = xy , while the total fencing needed is the sum of the perimeter 2 x +2 y and the inner partitions 2 y . Since 48 yards of fencing available we get a relation 48 = 2 x + 4 y i.e. , 24 = x + 2 y . Eliminating y from these two equations gives an expression A = 1 2 x (24 x ) = 12 x 1 2 x 2 for the area solely as a function of x . As the maximum value of x is x = 24, the maximum area will thus be the absolute maximum value of A on the interval [0 , 24]. This maximum will occur at a critical point x of A in (0 , 24) or at an endpoint. Now A ( x ) = 12 x = 0 when x = 12. But A (0) = 0 , A ( x ) = 72 , A (24) = 0 . Thus the maximum area of the dog pound is max area = 72 sq.yards and this occurs when the pound is 12 yards wide....
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 Fall '08
 schultz
 Calculus

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