At what points is the direction of fastest change of the function f(x,y)=x2+y2−4x−8y is i+j.

The required course is established by  ${\displaystyle \mathrm{\nabla }f(x,y)}$
${\displaystyle \mathrm{\nabla }f(x,y)=<\frac{\partial f}{\partial x},\frac{\partial f}{\partial }y)>}$
${\displaystyle \frac{\partial f}{\partial x}=\frac{\partial }{\partial x}(x^2+y^2-4x-8y)}$
$=2x-4$
${\displaystyle \frac{\partial f}{\partial y}=\frac{\partial }{\partial y}(x^2+y^2-4x-8y)}$
$=2y-8$
Thus,
${\displaystyle \mathrm{\nabla }f(x,y)=<\frac{\partial f}{\partial x},\frac{\partial f}{\partial }y)>}$
${\displaystyle =<2x-4,2y-8>}$
gradf(x,y) has to be parallel to i+j
The direction vector of i+j is ${\displaystyle <1,1>}$
Thus,
${\displaystyle <2x-4,2y-8>}$ has to be parallel to <1, 1>
Since in ${\displaystyle <1,1>}$ the coordinates are equal, 2x-4=2y-8
$2x-4=2y-8$
$2(x-2)=2(y-4)$
$x-2=y-4$
$y=x+2$