K EXAM 3-solutions - Version 013 – K EXAM 3 – ben-zvi –(55600 1 This print-out should have 18 questions Multiple-choice questions may

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 of A in (0 , 32) or at an endpoint. Now A ( x ) = 16- x = 0 when x = 16. But A (0) = 0 , A ( x ) = 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 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 , 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....
View Full Document

This note was uploaded on 12/05/2011 for the course M 408k taught by Professor Schultz during the Spring '08 term at University of Texas at Austin.

Page1 / 10

K EXAM 3-solutions - Version 013 – K EXAM 3 – ben-zvi –(55600 1 This print-out should have 18 questions Multiple-choice questions may

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online