# Find minimum of function $\frac{\left| x-12\right| }{5}+\frac{\sqrt{x^2+25}}{3}$

Find minimum of function $$f(x) = \frac{\left| x-12\right| }{5}+\frac{\sqrt{x^2+25}}{3}$$

I tried compute min using by definition of abs function. I consider two cases:

1. when $$x > 12$$ we have: $$f_{1}(x) = \frac{x-12}{5}+\frac{\sqrt{x^2+25}}{3}$$ using standard method $$f_{1}'(x)=0$$ for $$x_{0}=-15/4$$ but $$x_{0}$$ is not in $$[12, \infty]$$
2. when $$x \leq 12$$ we have $$f_{2}(x) = \frac{-x+12}{5}+\frac{\sqrt{x^2+25}}{3}$$ similarly $$f_{2}'(x) = 0$$, and $$x = 15/4$$

So using observation from this method I computed minimum of $$f(15/4) = 56/15$$.

Does it be a correct way?

• Your method appears to be correct. Aug 27, 2019 at 22:05
• 1) can be reasoned more simply that $f_1(x)$ is strictly increasing on $x > 12$ so there can't be any minimum there. Aug 27, 2019 at 23:23
• For good measure, I would also compute the value at the boundary $x=12$ to see how it compares. To see why this is necessary, try using your method to find the minimum of the function $f(x)=|x|$. Aug 28, 2019 at 0:55

$$\frac{|x-12|}{5}+\frac{\sqrt{x^2+25}}{3}=\frac{|x-12|}{5}+\frac{\sqrt{(3^2+4^2)(x^2+5^2)}}{15}\geq$$ $$\geq \frac{|x-12|}{5}+\frac{|3x+20|}{15}=\left|\frac{12}{5}-\frac{x}{5}\right|+\left|\frac{4}{3}+\frac{x}{5}\right|\geq$$ $$\geq \left|\frac{12}{5}-\frac{x}{5}+\frac{4}{3}+\frac{x}{5}\right|=\frac{56}{15}.$$ The equality occurs for $$(3,4)||(x,5)$$ and $$\left(\frac{12}{5}-\frac{x}{5}\right)\left(\frac{4}{3}+\frac{x}{5}\right)\geq0,$$ which gives $$x=\frac{15}{4},$$
• $\dfrac{|x-12|}5\ge0$ and $\dfrac{\sqrt{x^2+25}}3\ge5,$ but that does not mean the minimum of their sum is ever as low as $0+5$ Aug 27, 2019 at 22:10