Exam3_2 5 - 3 x y = 1 we have to minimize the function D x = x 2(1-3 x 2 1 2 =(10 x 2-6 x 1 1 2 on the interval £ 1 3 ±rom the graph it is clear

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Taylor, Douglas – Exam 3 – Due: Dec 5 2007, 1:00 am – Inst: JEGilbert 5 The critical points of A are thus the solutions of A 0 ( x ) = 14 - 2 x = 0 , i.e. x = 7. On practical grounds (or use the fact that A 00 = - 2), this will yield a maximum value of A ( x ). At x = 7, therefore, the enclosed area will have a maximum value A max = 729 sq. ft. keywords: optimization, fence, maximum, area 010 (part 1 of 1) 10 points If P ( x, y ) is a point on the line 3 x + y = 1 as shown in P ( x, y ) 1 1 3 (not drawn to scale), Fnd the value of x when P is closest to the origin. 1. x = 0 2. x = 2 5 3. x = 1 10 4. x = 3 10 correct 5. x = 1 5 Explanation: The distance from P ( x, y ) to the origin is given by D = p x 2 + y 2 , so when P lies on the line
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Unformatted text preview: 3 x + y = 1 , we have to minimize the function D ( x ) = ( x 2 + (1-3 x ) 2 ) 1 / 2 = (10 x 2-6 x + 1) 1 / 2 on the interval £ , 1 3 / . ±rom the graph it is clear that this minimum value occurs at a critical point of D in ( , 1 3 ) . Now, at a critical point of D , dD dx = 1 2 ‡ 20 x-6 (10 x 2-x + 1) 1 / 2 · = 10 x-3 (10 x 2-x + 1) 1 / 2 = 0 . Consequently, the P is closest to the origin when x = 3 10 . keywords: optimization, distance, minimum, point 011 (part 1 of 1) 10 points The rectangle in the Fgure ( x, y )...
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This note was uploaded on 02/14/2012 for the course MATH 408 K taught by Professor Clark,c.w./hoy,r.r during the Spring '08 term at University of Texas at Austin.

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