<s>[INST] Need help finding smallest value of $x^2 + y^2$ I need to find the smallest value of $x^2 + y^2$ with the restriction $2x + 3y = 6$. This chapter focuses on the vertex formula. 
 [/INST] For fun, we give a couple of solutions that are not the intended ones. The solutions are very similar, but the first is expressed algebraically, while the
second brings in the geometry.
$1$) Note that 
$$(2x+3y)^2+(3x-2y)^2=13(x^2+y^2).$$
Thus, given that $2x+3y=6$, 
$$13(x^2+y^2)= 36+(3x-2y)^2.$$
If we can manage to make $3x-2y=0$, then $13(x^2+y^2)$ will be as small as possible. But the system of two linear equations $2x+3y=6$, $3x-2y=0$ has a solution.  There,
$$13(x^2+y^2)=36,$$
so the smallest possible value of $x^2+y^2$ is $36/13$. 
$2$)  Look at the problem geometrically. We want to find the smallest radius $r$ such that the circle $x^2+y^2=r^2$ meets the line $2x+3y=6$. If we draw a picture, we can see that for this smallest $r$, the line $2x+3y=6$ will be tangent to the circle.  Let the point of tangency be $T(a,b)$.  The line from the origin to $T$ is perpendicular to the tangent line.
The line $2x+3y=6$ has slope $-2/3$.  So the line from the origin to $T$ has slope the negative of the reciprocal of $-2/3$. Thus
$$\frac{b}{a}=\frac{3}{2}.$$
This equation simplifies to $3a-2b=0$. We also have $2a+3b=6$. Now we can solve for $a$ and $b$. But let's not bother. Use the fact that
$$(3a-2b)^2+(2a+3b)^2=13(a^2+b^2)$$
to conclude that $r^2=a^2+b^2=36/13$.
</s>