Maximising and minimising $f(x,y)$ on $x^2+y^2\leqslant 9$ Find the absolute minimum and maximum of $f(x,y):=x^2+y^2-8y+3$ on the
  disc $x^2+y^2\leqslant 9$.


I know what the answer is and how to obtain it. 
What I do not understand is why we may assume that the extrema occur on the boundary of the disc.

I first looked at $Jf(x,y)=\mathbf{0}$ which occurs iff $(x,y)=(0,4)$, which is outside the disc.

One can then use the Lagrange Multiplier Theorem with the constraint $x^2+y^2=9$ to obtain the correct answers.
But I do not understand how we know this must be the constraint. How can we know for certain that the minimum and maximum do not occur in the interior of the disc?
You can  use a very elementary argument to argue that maximum and minimum must occur on the boundary. Suppose, if possible, the maximum occurs at a point $(a,b)$ with $a^{2}+b^{2} &lt;9$. By increasing/decreasing $a$ slightly you can  stay within the disk but make the first term in $f$ larger. This contradiction shows that maximum occurs on the boundary. Similarly for the minimum.
Since $y\geq-3$, we obtain:
$$x^2+y^2-8y+3\leq9-8\cdot(-3)+3=36.$$
The equality occurs for $x=0$ and $y=-3,$ which says that we got a maximal value.

Also, since $y\leq3$, we obtain:
$$x^2+y^2-8y+3=y^2-8y+3=-12+x^2+(3-y)(5-y)\geq-12.$$
The equality occurs for $x=0$ and $y=3$, which says that we got a minimal value.