This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 3 x + y = 1 , we have to minimize the function D ( x ) = ( x 2 + (13 x ) 2 ) 1 / 2 = (10 x 26 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 x6 (10 x 2x + 1) 1 / 2 · = 10 x3 (10 x 2x + 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 )...
View
Full
Document
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.
 Spring '08
 CLARK,C.W./HOY,R.R
 Critical Point

Click to edit the document details