$d \= \sqrt{x^{2 }\+{10}^2}$, $\mathrm{D}\= \sqrt{{\left(30-x\right)}^2 \+ {20}^2}$$$

$x \= \left\{ 0\le x \le 30\, x \in \mathrm{\ℝ}\right\}$

Total distance $\mathit{T}$:

$T$

$\=$

$d\+ \mathrm{D}$

$\=$

$\sqrt{x^2\+{10}^2 }\+ \sqrt{{\left(30-x\right)}^2\+{20}^2}$

Calculate first derivative:

$\frac{\ⅆT}{\ⅆ x}$

$\=$

$\frac{x}{\sqrt{x^2\+{10}^2}}- \frac{\left(30-x\right)}{\sqrt{{\left(30-x\right)}^2\+{20}^2}}$

Set it to 0:

$0$

$\=$

$\frac{x}{\sqrt{x^2\+{10}^2}}- \frac{\left(30-x\right)}{\sqrt{{\left(30-x\right)}^2\+{20}^2}}$

Solve for an x value that minimizes T:

$0$

$\=$

$x \sqrt{{\left(30-x\right)}^2\+{20}^2}- \left(30-x\right)\sqrt{x^2\+{10}^2}$

$\left(30-x\right)\sqrt{x^2\+{10}^2}$

$\=$

$x \sqrt{{\left(30-x\right)}^2\+{20}^2}$

${\left(30-x\right)}^2{\left(\sqrt{x^2\+{10}^2}\right)}^2$

$\=$

$x^{2 }{\left(\sqrt{{\left(30-x\right)}^2\+{20}^2}\right)}^2$

${\left(30-x\right)}^2\left(x^2\+{10}^2\right)$

$\=$

$x^{2 }\left({\left(30-x\right)}^2\+{20}^2\right)$

$\left(x^{2 } - 60 x \+ 900\right)\left(x^2\+100\right)$$$$$

$\=$

$x^{2 }\left(x^{2 } - 60 x \+ 900\+400\right)$

$x^4-60\mathbf{}{x}^3\+1000\mathbf{}{x}^2-6000 x\+90000$

$\=$

$x^{4 } - 60 x^3 \+ 1300 x^2$

$-300\mathbf{}{x}^2-6000 x \+90000$

$\=$

$0$

$-300\mathbf{}\left(x\+30\right)\mathbf{}\left(x-10\right)$

$\=$

$0$

Choose the positive root:

$x$

$\=$

$10$