A ship lies 6 miles from a straight shore, and opposite a point 10 miles farther along the shore, another ship lies 18 miles offshore.

By Pythagorean Theorem
 

 

$d_{min} = \sqrt{10^2 + (6 + 18)^2}$

$d_{min} = 26 ~ \text{mi}$

$d_1 = \sqrt{x^2 + 6^2}$

$d_1 = \sqrt{x^2 + 36}$
 

$d_2 = \sqrt{(10 - x)^2 + 18^2}$

$d_2 = \sqrt{(100 - 20x + x^2) + 324}$

$d_2 = \sqrt{424 - 20x + x^2}$
 

 

The distance the boat can travel
$d = d_1 + d_2$

$d = \sqrt{x^2 + 36} + \sqrt{424 - 20x + x^2}$
 

For the least distance the boat can travel
$\dfrac{dd}{dx} = \dfrac{2x}{2\sqrt{x^2 + 36}} + \dfrac{-20 + 2x}{2\sqrt{424 - 20x + x^2}} = 0$

$\dfrac{-20 + 2x}{2\sqrt{424 - 20x + x^2}} = \dfrac{2x}{2\sqrt{x^2 + 36}}$

$\dfrac{x - 10}{\sqrt{424 - 20x + x^2}} = \dfrac{x}{\sqrt{x^2 + 36}}$

$(x - 10)\sqrt{x^2 + 36} = x\sqrt{424 - 20x + x^2}$

$(x - 10)^2 (x^2 + 36) = x^2 (424 - 20x + x^2)$

$(x^2 - 20x + 100)(x^2 + 36) = 424x^2 - 20x^3 + x^4$

$x^4 + 36x^2 - 20x^3 - 720x + 100x^2 + 3600 = 424x^2 - 20x^3 + x^4$

$36x^2 - 720x + 100x^2 + 3600 = 424x^2$

$288x^2 + 720x - 3600 = 0$

$x = 2.5 ~ \text{mi}$
 

The least distance
$d = \sqrt{2.5^2 + 36} + \sqrt{424 - 20(2.5) + 2.5^2}$

$d_{min} = 26 ~ \text{mi}$