{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

K EXAM 3-solutions - Version 013 K EXAM 3 ben-zvi(55600...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Version 013 – K EXAM 3 – ben-zvi – (55600) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A rectangular dog pound with three kennels as shown in the figure consists of a rectangular fenced area divided by two partitions. Determine the maximum possible area of this pound if 64 yards of chain link fencing is available for its construction. 1. max area = 130 sq.yards 2. max area = 129 sq.yards 3. max area = 131 sq.yards 4. max area = 128 sq.yards correct 5. max area = 132 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 64 yards of fencing available we get a relation 64 = 2 x + 4 y i.e. , 32 = x + 2 y . Eliminating y from these two equations gives an expression A = 1 2 x (32 - x ) = 16 x - 1 2 x 2 for the area solely as a function of x . As the maximum value of x is x = 32, the maximum area will thus be the absolute maximum value of A on the interval [0 , 32]. This maximum will occur at a critical point x 0 of A in (0 , 32) or at an endpoint. Now A ( x ) = 16 - x = 0 when x 0 = 16. But A (0) = 0 , A ( x 0 ) = 128 , A (32) = 0 . Thus the maximum area of the dog pound is max area = 128 sq.yards and this occurs when the pound is 16 yards wide. 002 10.0 points If P ( x, y ) is a point on the line 2 x + y = 5 as shown in P ( x, y ) 5 5 2
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Version 013 – K EXAM 3 – ben-zvi – (55600) 2 (not drawn to scale), find the value of x when P is closest to the origin. 1. x = 14 5 2. x = 12 5 3. x = 13 5 4. x = 2 correct 5. x = 11 5 Explanation: The distance from P ( x, y ) to the origin is given by D = x 2 + y 2 , so when P lies on the line 2 x + y = 5 , we have to minimize the function D ( x ) = ( x 2 + (5 - 2 x ) 2 ) 1 / 2 = (5 x 2 - 20 x + 25) 1 / 2 on the interval 0 , 5 2 . To locate the critical points of D we solve the equation dD dx = 1 2 10 x - 20 (5 x 2 - x + 25) 1 / 2 = 5 x - 10 (5 x 2 - x + 25) 1 / 2 = 0 , i.e. , x = 2 . Since this critical point lies in (0 , 5 2 ), we see that D ( x ) will have a mini- mum value when x = 2 by simply looking at the figure, but we can also show this using the First Derivative Test and the sign chart 2 - + for dD/dx . Consequently, the point P is closest to the origin when x = 2 . 003 10.0 points A homeowner wants to build a fence to enclose a 180 square yard rectangular area in his backyard. Along one side the fence is to be made of heavy-duty material costing $9 per yard, while the material along the remaining three sides costs $1 per yard. Determine the least cost to the homeowner. 1. least cost = $110 2. least cost = $115 3. least cost = $120 correct 4. least cost = $130 5. least cost = $125 Explanation: Let x be the length of the side made of the heavy-duty material and y the length of an adjacent side. Then we want to minimize the cost function C ( x, y ) = 10 x + 2 y ,
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern