<s>[INST] If $ 5x+12y=60$ , what is the minimum of $\sqrt{x^2+y^2}$? I know this can be easily done by solving for $y$ and substituting, so that you only have to find the minimum value of the parabola $\large x^2 + \left ( \frac {60-5x}{12} \right)^2$ using standard techniques, but is there a less messy way to do this using inequalities? I tried various things such as $AM-GM$, but I don't get anywhere. One thing which I did notice about the restriction is that it is of the form $ax+by=ab$, but I don't know how to make any use of that. Thanks.
 [/INST] You need to find the foot of the perpendicular to the line $5x + 12y = 60$ that passes through the origin.  The line has slope $-\frac{5}{12}$, so any perpendicular has slope $\frac{12}{5}$.  The perpendicular passes through the origin, so its equation is $y=\frac{12}{5}x$.  Find the intersection of the two lines:
\begin{aligned}
5x + 12y &= 60\\
y &= \frac{12}{5}x\\
5 x + \frac{12 \cdot 12}{5} x &= 60\\
25x + 144 x &= 300\\
169x &= 300\\
x &= \frac{300}{169}\\
y &= \frac{12}{5} \cdot \frac{300}{169}\\
y &= \frac{720}{169}.
\end{aligned}
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