2 posts one is 10m high and other is 15m higs stand 30m apart they are to be stayed by transmission wire attached to a single stake at the ground level, the wire running to the top of the post , where should stake place to used least amount of wire ?
In the above solution, the actual calculation consists of the last 4 lines, the rest explains why it works
otherwise .... CALCULUS
AP^2 = x^2 + 10^2
AP = √(x^2 + 100)
PD^2 = (30-x)^2 + 15^2
PD = √(x^2 - 60x + 1125)
L = √(x^2 + 100) + √(x^2 - 60x + 1125)
= (x^2 + 100)^(1/2) + (x^2 - 60x + 1125)^(1/2)
dL/dx = (1/2)(x^2 + 100)^(-1/2) (2x) + (1/2)(x^2 - 60x + 1125)^(-1/2) (5x - 60)
= 0 for a min of L
x/√(x^2 + 100) = -(x-30)/√(x^2 - 60x + 1125)
square both sides
x^2/(x^2 + 100) = (x^2 - 60x + 900)/(x^2 - 60x + 1125)
cross multiply
x^4 - 60x^3 + 900x^2 + 100x^2 - 6000x + 90000 = x^4 - 60x^3 + 1125x^2
125x^2 + 6000x - 90000=0
x^2 + 48x - 720 = 0
(x-12)(x+60) = 0
x = 12 or x = -60 , but x can't be negative, so
x = 12
simplest way:
draw a parallelogram ABCD, where AB = 10 (first post)
CD = 15 (2nd post) and BC = 30
Reflect AB in the line BC to point A1 , so that AB= BA1
joint DA1 intersecting BC at P
length of wire = AP+DP = A1P+DP , which is a straight line, and the shortest distance between two points is a straight line, ahhhhh....
let BP=x
Triangles ABP and A1PB are congruent, thus AP=A1P.
Also angles BPA1 and CPD are opposite, thus equal, making triangles APB and DPC similar
Therefore:
x/10 = (30-x)/15
15x = 300-10x
25x=300
x = 12
Put stake 12 m from the 1st post
[ ( Short post height ) x ( distance ) ] all divide [ short post height + large post height ]
[ ( 10 ) x ( 30 ) ] [300]
———————. —>. ——- --> 12
[ 10 + 15 ]. [ 25 ]
Above one mixed up between the spaces last step is 300/25 =12
Well, if you want to use the least amount of wire, I would suggest placing the stake right in the middle, at a distance of 15 meters from each post. That way, you'll have equal lengths of wire running from the stake to the top of each post. Not only will it save you wire, but it'll also keep things balanced, just like a good joke!
To minimize the amount of wire needed, the stake should be placed at the point that divides the total distance between the two posts proportionally to their respective heights. Let's calculate this using a concept called similarity of triangles.
1. First, let's assign some variables:
- Let h1 be the height of the first post (10m).
- Let h2 be the height of the second post (15m).
- Let d be the distance between the two posts (30m).
- Let x be the distance from the stake to the first post.
2. We can set up a proportion based on the similarity of triangles formed:
- (x / h1) = ((d - x) / h2)
3. Next, let's solve this proportion for x:
- Cross-multiply: x * h2 = (d - x) * h1
- Expand: x * h2 = d * h1 - x * h1
- Rearrange: x * h2 + x * h1 = d * h1
- Combine like terms: x * (h2 + h1) = d * h1
- Solve for x: x = (d * h1) / (h2 + h1)
4. Substitute the given values into the formula:
- x = (30 * 10) / (15 + 10)
- x = 300 / 25
- x = 12
Therefore, the stake should be placed 12 meters away from the first post and 18 meters away from the second post to minimize the amount of wire needed.