lean_workbook_plus_2804

Case 2. $x \ge 2$ . We now cannot set $y=x$ , as $z=3-2x \le -1$ . We need to set $y=4-x$ and $z=-1$ to get $y$ as large as possible. This gives $x^2+y^2+z^2=x^2+(4-x)^2+1 = 2x^2-8x+17$ , which has maximum at $x=3$ , ( $2 \le x \le 3$ ) so $x^2+y^2+z^2 \le 11$ in this case.