**Problem 01**

A 5-m line AD intersect at 90° to line BC at D so that BD is 2 m and DC = 3 m. Point P is located somewhere on AD. The total length of the cables linking P to points A, B, and C is minimized. How far is P from A?

**Solution 01**

$L = L_1 + L_2 + L_3$

$L = \sqrt{2^2 + y^2} + \sqrt{3^2 + y^2} + (5 - y)$

$L = \sqrt{4 + y^2} + \sqrt{9 + y^2} + 5 - y$

$\dfrac{dL}{dy} = \dfrac{2y}{2\sqrt{4+y^2}} + \dfrac{2y}{2\sqrt{9+y^2}} - 1 = 0$

$y = 1.407 ~ \text{m}$

$L_3 = 5 - y$

$L_3 = 5 - 1.407$

$L_3 = 3.593 ~ \text{m}$ *answer*