Answer
a) $4x+4y+4$ b) $10x+10y-5$
Work Step by Step
a) Since, we have $L(x,y)=f(x,y)+f_x(x-a)+f_y(y-b)$ ...(1)
Now,
$f_x=2x+2y+2z \implies f_x(0,0)=4$; and $f_y=2x+2y+2z \implies f_y(0,0)=4$
and $f(0,0)=4$
Now eq. (1) becomes $L(0,0)=4+4(x-0)+4(y-0)=4x+4y+4$
b) Since, we have $L(x,y)=f(x,y)+f_x(x-a)+f_y(y-b)$ ...(1)
Now,
$f_x=2x+2y+2z \implies f_x(1,2)=10$; and $f_y=2x+2y+2z \implies f_y(1,2)=10, f(1,2)=25$
Now eq. (1) becomes $L(1,2)=25+10(x-1)+10(y-1)=10x+10y-5$
Hence, a) $4x+4y+4$ b) $10x+10y-5$
