<s>[INST] Help with minimization problem help me, if $x$ and $y$ are real such that $3x-4y = 12$, determine the minimum value of $z = x ^ 2 + y ^ 2$?$$$$I thought of $$3x-4y = 12\Longrightarrow x=4\frac{y+3}{3}\\z = x ^ 2 + y ^ 2\Longrightarrow z = \left(4\frac{y+3}{3}\right) ^ 2 + y ^ 2$$ and then?
 [/INST] To answer your question, "and then?", with details, this is a way the calculation you began could continue, and this is without calculus:
$z=(4\frac{y+3}{3})^2 + y^2 = 16\frac{(y+3)^2}{9} + \frac{9y^2}{9} = \frac{16(y^2+6y+9)+9y^2}{9} = \frac{25}{9}y^2 + \frac{32}{3}y + 16$.
Center of a parabola is at $y = \frac{-b}{2a} = \frac{-32/3}{50/9} = -\frac{48}{25}$.
Finally, $x=4\frac{y+3}{3} = \frac{4(27/25)}{3}  =\frac{36}{25}$.
The method described by Brian Scott is a bit easier, I think.
</s>