Chapter 13 - Section 13.6 - Tangent Planes and Differentials - Exercises - Page 729: 34

$0.02$

$f_x= x+y+3 \implies f_x(2,2)=2+2+3=7$ and $f_y(x,y) =x+\dfrac{y}{2}-3 \implies f_{y}(2,2) =2+\dfrac{2}{2}-3=0$ $f_{xx}(x,y)=1; f_{yy}(x,y)=\dfrac{1}{2}$ and $f_{xy}(x,y) =1$ $L(x,y)=3+1(x-2) -6(y-1) =7+x-6y$ The error can be found as: $|E(x,y)| \leq \dfrac{1}{2} \times 1 [ |x-2| +|y-2|)^2$ $\implies E \leq \dfrac{1}{2} \times (0.1+0.1)^2 =0.02$