Diagonalize the quadratic forms in Exercises 3540 by finding an orthogonal matrix Q such that the change of variable x Qy transforms the given form into one with no cross-product terms. Give Q and the new quadratic form.

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 √ lnyy (1, 2) =0 lnxy (1, 2) =(2)( 2) = 2 2 2 H (1, 2)= 2 * 0 (2 2) 08= 8< 0 Since H is negative, we have a saddle point at (1, 2). Critical point at (1, + 2) : lnxx (1, + 2)=2 lnyy (1, + 2)=0 lnxy (1, + 2)= 2( 2) 2 H(1, + 2) = 2* 0 (2 2)= 08= 8<0 Since H is negative, we have a saddle point at (1, + 2). Homework #11: M(c,f)= 11+8c+4c 240f20cf+40f2 Mc= 8+8c20f> 8+8c20f= 0 divide by 4 2+2c5f=0 Mf= 4020c+80f> 4020c+80f=0 divide by 20 2c+4f=0