two poles 6m and 15m tall are 20 m apart a wire is attached to the top of each pole and is also staked to the ground some where b/n two poles where should be the wire be staked so that the minimum amount of wire be used?

make a sketch.

draw poles AB and PQ with B and Q on the ground
Draw the reflection of AB in the horizontal and call it BA'. Joint A' to P
Label the intersection of A'P with BQ as M.
let BM = x, then
MQ = 20-x

Of course the shortest distance between any two points is the straight line joining them, So A'P is the shortest
but A'P = AM + MP <--- the shortest

since triangles ABM and PQM are similar,
AB/BM = PQ/MQ
6/x = 15/(20-x)
15x = 120 - 6x
21x = 120
x = 120/21 = 40/7

Attach it 40/7 m from B

thanks it is the best very much but where is the graph of the triangles

Well, well, well, looks like we've got a real "wires" situation here! Let's figure it out, shall we? Now, imagine those poles are party guests, desperately trying to connect without causing a fuss. To minimize wire usage, we need to find the perfect spot for our wire to stake its claim.

First things first, we can visualize that the shortest wire path would be a straight-line connection between the tops of the poles, forming a right angle triangle. The 20-meter distance between the bases of the poles acts as the hypotenuse of this triangle.

Now, let's apply some mathematical magic! According to the Pythagorean Theorem, the square of the hypotenuse (20m) is equal to the sum of the squares of the other two sides (6m and 15m). So, substituting those values, we get the equation:

20^2 = 6^2 + 15^2

Simplifying that, we have:

400 = 36 + 225

This means that the square of the wire length is equal to 261. To find the actual length of the wire, we take the square root of 261, which is roughly 16.14 meters.

So, to minimize the wire usage, stake the wire approximately 16.14 meters away from either pole, forming a right angle with the wire and the ground. Voilà! The poles can now connect in the most wire-saving way possible!

To determine the location where the wire should be staked to minimize the amount of wire used, we can apply the concept of similar triangles.

First, let's consider the situation as illustrated below:

A
|\
| \
| \
6m \ 15m
| \
| \
|______\
B 20m C

In the diagram, A and B represent the tops of the poles, while C represents the location where the wire needs to be staked. We need to find the position of C that minimizes the length of the wire.

To begin, let's form two similar triangles: triangle ABC and triangle AED.

In triangle ABC:
AB represents the height of the shorter pole, which is 6m.
AC represents the height of the taller pole, which is 15m.
BC represents the distance between the poles, which is 20m.

In triangle AED:
AE represents the height of the shorter pole, which is 6m.
ED represents the height of the wire from the ground to the point where it is staked.
AD represents the height of the taller pole, which is 15m.

Now we can establish the ratios between the corresponding sides of both triangles:

AB/AC = AE/ED

Substituting the known values:
6m/15m = 6m/ED

Simplifying the equation:
6m * ED = 15m * 6m
ED = (15m * 6m) / 6m
ED = 15m

Therefore, the height from the ground to the point where the wire should be staked is 15m.

So, the wire should be staked at a height of 15 meters from the ground, which is the point directly below the top of the taller pole.

To determine where the wire should be staked in order to minimize the amount of wire used, we can use the concept of similar right triangles.

Let's consider the two poles as P1 and P2, with heights of 6m and 15m respectively. The distance between the poles is 20m.

First, we need to visualize the situation. Draw a diagram of the two poles and the wire attached to the top of each pole. Since the wire is anchored somewhere between the two poles, we need to find the optimal position for it.

Let's assume the wire is staked at a point X, at a distance d from pole P1 and (20 - d) from pole P2. We need to find the value of d that minimizes the total length of the wire.

Now, let's consider the triangles formed by the wire and the poles. We have two similar right triangles: P1XQ and P2XR, where Q and R are the points where the wire touches the ground.

Since the triangles are similar, we can set up the following proportion:

(P2R) / (P1Q) = (P2P1) / (P1X)

We know that P2P1 = 20m and P2R = 15m, so the equation becomes:

15 / (P1Q) = 20 / (P1X)

To minimize the length of the wire, we need to minimize (P1Q) + (P2R), which is equivalent to minimizing (P1Q) + 15.

Let's substitute (P1Q) with (20 - d) and solve for d:

15 / (20 - d) = 20 / (P1X)

Cross-multiplying, we get:

15 * (P1X) = 20 * (20 - d)

Simplifying, we have:

15 * (P1X) = 400 - 20d

Now, let's simplify further:

15 * (P1X) + 20d = 400

15 * (P1X) = 400 - 20d

Divide both sides by 5:

3 * (P1X) = 80 - 4d

Now, separate the variables:

3 * (P1X) = 80 - 4d

80 = 4d + 3 * (P1X)

80 - 4d = 3 * (P1X)

Divide both sides by 3:

(80 - 4d) / 3 = P1X

Finally, to minimize the amount of wire used, the wire should be staked at a point X, which is at a distance (80 - 4d) / 3 from pole P1.

By finding the value of d, you can substitute it into the equation to calculate the exact position for staking the wire.