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$$ and then?
 A: The quantity $x^2+y^2$ is minimized when $\sqrt{x^2+y^2}$ is minimized, and $\sqrt{x^2+y^2}$ is just the distance from the point $\langle x,y\rangle$ to the origin. Thus, you’re looking for the point on the line $3x-4y=12$ that is closest to the origin. Call this line $\ell$; the point closest to the origin is the point of intersection of $\ell$ with a line through the origin and perpendicular to $\ell$. The slope of $\ell$ is $\frac34$, so the slope of the perpendicular is $-\frac43$: you want the intersection of $\ell$ and the line $y=-\frac43x$.
A: We start from your expression for $z$ in terms of $y$. The standard "calculus" way of handling the problem is to calculate $\frac{dx}{dy}$, and set it equal to $0$ to find the critical points.
For the calculation of $\frac{dz}{dy}$, you can first simplify, and then differentiate, or differentiate, and then simplify. 
I will do it one way, and you can do it the other way. We have 
$$z=\frac{16}{9}(y+3)^2+y^2.$$
Differentiate. We get
$$\frac{dz}{dy}=\frac{32}{9}(y+3)+2y=\frac{1}{9}(50y+96).$$
Set this equal to $0$ and solve for $y$. 
Don't forget to check that we really do get a minimum.
A: Then it's a parabola, which I suspect you know how to minimize. (Or you can use calculus if you want to).
A: To answer your question, "and then?", with details, this is a way the calculation you began could continue, and this is without calculus:
$z=(4\frac{y+3}{3})^2 + y^2 = 16\frac{(y+3)^2}{9} + \frac{9y^2}{9} = \frac{16(y^2+6y+9)+9y^2}{9} = \frac{25}{9}y^2 + \frac{32}{3}y + 16$.
Center of a parabola is at $y = \frac{-b}{2a} = \frac{-32/3}{50/9} = -\frac{48}{25}$.
Finally, $x=4\frac{y+3}{3} = \frac{4(27/25)}{3}  =\frac{36}{25}$.
The method described by Brian Scott is a bit easier, I think.
A: $$
{\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}
$$
