Question

U Question 22 1 pts Find the absolute minimum of f(x, y) = x2 + 4y? - 2x²y + 4 on the square given by -1 << < 1 and -1<y<1. 11 4 8 None of the above or below O-2 07

Answer 1

you 7--1 (I) x=-1 4-1 -1975 22 so 2x-eny=0 Absolute Minimum of feniy)=x²+442_28124 +4 on the esquere given by tened and olsus! Now on the beundry of squest -14441 left side (I) so f(-44)=412+472-22-132y + 4 43-1 917)=f(+14) = 5+442-27 1) critical points critical point of gly) fx=0 7 21-447=0 gley)=0 7 qy-2= fy=0 84-2x2=0 so, 91-1) = 5+464)2-26-1) =1196-1) = 11 714= 914) = 5+4(+34-2() 9,16) =42475 912)5+4104-2(1) -1941):7 21- 40 x 20 2x-x3=0 9(-1)=11 F(-1,-1) =1112 212-72) = 0 914) = 4.75 = 1F 1-1,5) -4.757 220, 2-42-0 943)=7 = littlt)=7 112=182 outside of the on the right side (II) demain 2-) i -1345) -ICHEL f(0,9) = 72+442 +2 +3444 so, azo 2) 947) Lly)= 8+47²+2y 7920 (htical point Cutical pourt 110,0) h'ey)=0 = 84-2201974 So, Hooge 4 since, 250 youth

111) on the lower domani So, h(-1)2 461,-1) = 5791-04-2-1 -1711,-1)= || hl=1H*15) = 4.75 So He menimum h(t)=\f11,1)=7 of all the functional value over the domain of squre is 4: -4,-XEL 9160,0)=4 f(x,-1)=x²+4(-12-20²1-1)+. *** Kix) = 3x2 +8 Absolute minimum=14 cntical pour Kly) - 6x=0 second option is consect. 720 so, K10)-8 >110-1)=8 K(-1) = 31-137+8 ) K1-11-11 off(-1,-1)=1) KOJI-311+8=11 - F-1) an the oboke domain -15XEL -2x=0 ه لا ال= t( 91 +1) = 24+4 (02-2741374 L60) = x -x2+8 L'ias=0 7 Lon=8 /f(0,4)=8 L(-1) = H218 = L(-1)=7 111-1,41):7 L(&)=-11²+887 - (+(,1)=7