Answer:
20 ft from the left pole
Step-by-step explanation:
Let x (ft) be the distance from the base of the left pole to the anchor point. Since 2 poles has a distance of 30 ft, the distance from the anchor point to the right pole is 30 - x
The wire length from the top of the left pole to the anchor point is
[tex]L_l^2 = 20^2 + x^2 = x^2 + 400[/tex]
Similarly for the 2nd part of the wire
[tex]L_r^2 = 10^2 + (30 - x)^2 = 100 + 900 -60x + x^2 =x^2 + 1000 - 60x[/tex]
We want to minimize the length of the wires, or [tex]L_l + L_r[/tex]
[tex] \sqrt{x^2 + 400} + \sqrt{x^2 + 1000 - 60x}[/tex]
we can take the first derivative of this and set it to 0
[tex] \frac{x}{\sqrt{x^2 + 400}} + \frac{x - 30}{\sqrt{x^2 + 1000 - 60x}} = 0[/tex]
x = 20
