f(x,y)=3x^2+y^2-3xy+6x-4y 求極小值=?
2006-06-14 10:09:08 · 2 個解答 · 發問者 ? 1 in 科學 ➔ 數學
方法一:f(x,y)=3x2+y2-3xy+6x-4y 對x偏微分f'(x,y)x=6x-3y+6對y偏微分f'(x,y)y=2y-3x-46x-3y+6=0 , 2y-3x-4=0 ==> x=0 , y=2==> f(x,y)=3x2+y2-3xy+6x-4y 在x=0 , y=2時有極小值 -4方法二:f(x,y)=3x2+y2-3xy+6x-4y =(y-(3/2)x-2)2+(3/4)x2-4在x=0 , y=2時有極小值 -4
2006-06-14 10:23:30 · answer #1 · answered by popo 6 · 0⤊ 0⤋
f(x,y)=3x2+y2-3xy+6x-4yfx(x,y)=6x-3y+6fy(x,y)=2y-3x-4fxx(x,y)=6fyy(x,y)=2fxy(x,y)=-3fyx(x,y)=-3令fx(x,y)=fy(x,y)=0=>6x-3y+6=02y-3x-4=0解得:(x,y)=(0,2)fxx(0,2)>0,fxx(0,2)*fyy(0,2)-(fxy(0,2))2=12-9>0∴f(x,y)在(0,2)有相對極小值,其值=f(0,2)=-4
2006-06-14 10:29:22 · answer #2 · answered by ? 7 · 0⤊ 0⤋