Given $3x+4y=15$, $\min(\sqrt{x^2+y^2})=?$ (looking for other approaches) 
Given, $(x,y)$ follow $3x+4y=15$. Minimize $\sqrt{x^2+y^2}$.

I solved this problem as follows,
We have $y=\dfrac{15-3x}{4}$,
$$\sqrt{x^2+y^2}=\sqrt{x^2+\frac{(3x-15)^2}{16}}=\frac{\sqrt{25x^2-90x+225}}4=\frac{\sqrt{(5x-9)^2+144}}{4}$$Hence $\min(\sqrt{x^2+y^2})=3$.
I'm wondering is it possible to solve this problem differently?

Go with Trigonometry.
Let $x=a\sin \alpha ,\, y=a\cos \alpha$, then we have:
$$\begin{aligned}&3a\sin \alpha+4a\cos \alpha=15\\
\implies &15\leq \sqrt {9a^2+16a^2}=5|a|\\
\implies &|a|\ge 3.\end{aligned}$$
$$\sqrt {x^2+y^2}=|a|\ge 3.$$
