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?

$$
{\rm F} \equiv x^{2} + y^{2} -\mu\left(3x - 4y - 12\right)\,,
\quad
\left\vert
\begin{array}{rcl}
{\partial F \over \partial x} = 0
&\Longrightarrow&
2x - 3\mu = 0\ \Longrightarrow\ x = {3 \over 2}\,\mu
\\[1mm] 
{\partial F \over \partial y} = 0
&\Longrightarrow&
2y + 4\mu = 0\ \Longrightarrow\ y = -2\mu
\end{array}\right. 
$$
$$
12 = 3x - 4y = {9 \over 2}\,\mu + 8\mu = {25 \over 2}\,\mu\ \Longrightarrow\
\mu = {24 \over 25}
\quad\Longrightarrow\quad
\left\vert%
\begin{array}{rcl}
x & = &{3 \over 2}\,{24 \over 25} = {36 \over 25}
\\[2mm]
y & = & -2\,{24 \over 25} = -\,{48 \over 25}
\end{array}\right.
$$
$$
\begin{array}{|c|}\hline\\
{\large\quad%
z_{\rm min}
=
\left(36 \over 25\right)^{2} + \left(-\,{48 \over 25}\right)^{2}
=
\color{#0000ff}{144}\quad}
\\ \\ \hline
\end{array}
$$
