y 2 sec x 61

MAT 211 Lecture 6 15.3 & 15.4 Section 15.3 Homework #8: ln(x,y)= x +y2yx2 lnx= 0 & lny= 0 1st Derivative: lnx= 2xy lny= 2y2yx I. 2xy=0 > x=1 2y=0 2=y y= 2 2 2y2yx=0 > y=0 2xy =0 2x0=0 x=0 II. 2y(1x)= 0 y=0 & x=1 Critical Points: (0,0); (1, 2); (1, + 2) 2nd Derivative: lnxx= 2 lnxx (0,0)= 2 lnyy= 22x lnyy (0,0)= 2 lnxy= 2y lnxy (0,0)= 0 H(0,0)= 2 * 2 0= 4 Since lnxx= 2>0, then we have a relative minimum at (0,0). Critical point at (1, 2) : lnxx (1, 2) =2 √