Homework 11-solutions

# Homework 11-solutions - wtm369 – Homework 11 –...

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Unformatted text preview: wtm369 – Homework 11 – Helleloid – (58250) 1 This print-out should have 14 questions. Multiple-choice 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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Homework 11-solutions - wtm369 – Homework 11 –...

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