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.