Question

Two poles are connected by a wire that is also connected to the ground. The first pole is 20 ft tall and the second pole is 10 ft tall. There is a distance of 30 ft between the two poles. Where should the wire be anchored to the ground to minimize the amount of wire need

Answer 1

Answer:

Therefore the wire should be anchored at 10 ft away from pole which is 10 ft long.

Step-by-step explanation:

Given that , The distance between two poles is 30 ft.

The length of 1st pole is = 20 ft

The length of second pole is = 10 ft.

Let the wire anchored to the ground at a distance x ft from the second pole.

Then, the distance of anchored from the first pole is = (30-x)

The total length of the wire is L = m+n

We know the pythagorean theorem,

Height²+base² = hypotenuse²

To find the value of m and n we use  pythagorean theorem

From the left side triangle in the picture we get,

10²+x²= m²

⇒m²=100+x²

$\Rightarrow m= \sqrt {100+x^2$

and right side  triangle in the picture we get,

20²+(30-x)² = n²

⇒n²= x²-60x+1300

$\Rightarrow n= \sqrt {x^2 -60x+1300}$

Then ,

$L= \sqrt{(100+x^2)}+\sqrt{(x^2-60x+1300) }$

Differentiating with respect to x

$L'= \frac {2x}{2\sqrt{100+x^2}}+ \frac{2x-60}{2\sqrt {x^2-60x+1300}}$

For minimize, L' =0

$\frac {2x}{2\sqrt{100+x^2}}+ \frac{2x-60}{2\sqrt {x^2-60x+1300}}=0$

$\Rightarrow \frac {x}{\sqrt{100+x^2}}=- \frac{x-30}{\sqrt {x^2-60x+1300}}$

Squaring both sides

$\Rightarrow( \frac {x}{\sqrt{100+x^2}})^2=(- \frac{x-30}{\sqrt {x^2-60x+1300}})^2$

$\Rightarrow x^2(x^2-60x+1300)= (x^2-60x+900)(100+x^2)$

$\Rightarrow x^4 -60x^3+1300x^2= 100x^2-6000x+90000+x^4-60x^3+900x^2$

$\Rightarrow 300x^2+6000x-90000=0$

$\Rightarrow x^2+20x-300=0$

$\Rightarrow x=10,-30$

Therefore x = 10. [x=-30 negligible, since distance can not negative]

Therefore the wire should be anchored at 10 ft away from pole which is 10 ft long.