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.
 A: Hint: Consider the graph of the line $5x+12y = 60$.
Hint: The minimum value occurs when the circle with origin as center, is tangential to the above line.
Hint: Find the point of tangency. Use the fact that a tangent is perpendicular to the radius.

Apply (weighted) QM-AM, we have
$\sqrt{\frac{x^2 + y^2}{169}} = \sqrt{\frac{25 \times \frac{x^2}{25} + 144 \times \frac{y^2}{144}}{169}} \geq \frac{ 25 \times \frac{x}{5} + 144 \times \frac{y}{12} } { 169 } = \frac{ 5x+12y}{169} = \frac{60}{169}$.
SO $\sqrt{x^2+y^2} \geq \frac{60}{13} $
A: Note that
$$(5x+12y)^2+(12x-5y)^2=(13^2)(x^2+y^2).$$
Since $5x+12y$ is given, we minimize $x^2+y^2$ by minimizing $(12x-5y)^2$. The minimum value of this is $0$. It follows that the minimum value of $\sqrt{x^2+y^2}$ is $\dfrac{60}{13}$. 
A: You can find the minimum distance from point $(0,0)$ to the mentioned straightline finding the perpendicular straightline which passes through $(0,0)$ and then its intersection.
A: You can draw a graph of $5x+12y=60$. $\sqrt{x^2+y^2}$ is minimum when it is the shortest distance from origin to $5x+12y=60$. The distance can be found using the area of the triangle and length of the hypotenuse. Area = $30$, hypotenuse = $13$. Therefore, distance = $\dfrac{60}{13}$
A: We have $\langle z,v\rangle = 60$ for $z:=(x,y)$ and $v=(5,12)$. We want to minimize $|z|$, or what is the same $|z|^2=\langle z,z\rangle$. Now, from Cauchy's inequality $|\langle z,v\rangle|\leq |z||v|$. From there you get the bound $|\langle z,v\rangle|/|v|\leq |z|$. The equality in Cauchy's inequality is attained when the vectors are proportional. So you get the minimum is$|\langle z,v\rangle |/|v|$.
A: $$5x+12y=60\implies\;\;(**)\;\; y=-\frac5{12}x+5\implies x^2+y^2=\frac{169}{144}x^2-\frac{50}{12}x+25$$
Since the minimum/maximum of $\,\sqrt{x^2+y^2}\,$ is the same as $\,x^2+y^2\;$ (why?).
Differentiate the above to get critical points:
$$\frac{169}{72}x-\frac{50}{12}=0\implies x=\frac{300}{169}\stackrel{(**)}\implies y=\frac{720}{169}$$
So the minimal value is
$$\sqrt{\frac{300^2+720^2}{169^2}}=\frac{780}{169}=\frac{60}{13}$$
A: 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}
A: One can use the Cauchy Schwarz Inequality, by which $(x^2+y^2)(25+144) \geq (5x+12y)^2$. Therefore, we see that the minimum value of $\sqrt{x^2+y^2}$ is $\sqrt{\frac{60^2}{13^2}}$, or $\boxed{\frac{60}{13}}$
A: I did this with similar triangles:
The constraint is a 5-12-13 right triangle (in Quadrant I), and the objective is simply the Euclidean distance of $(x,y)$ to the origin.
Since we're minimizing the distance from a point (the origin) to a line, it should be clear from the graph we only need to focus on Quadrant I.
Also, to minimize the distance from a point to a line, connect the point to the line with a segment perpendicular to the line.
If you've been drawing, we now have a similar triangle within the 5-12-13 triangle.

Our minimum distance is given by:
$\frac{12}{13}=\frac{x}{5}\Rightarrow x=\frac{60}{13}$
