# Thread:

1. ## The bird and the seed

There is this bird sitting on the top of the bar with length "L2".
The bird has to come down, pick a seed from the farm and go and sit on the top of the bar with length "L1".

The distance between the two bars is "S"

Propose an equation using which the bird will pass MINIMUM way to do this operation.
(Assume L1 is not equal to L2)

2. ## Re: The bird and the seed

Originally Posted by Narek
There is this bird sitting on the top of the bar with length "L2".
The bird has to come down, pick a seed from the farm and go and sit on the top of the bar with length "L1".

The distance between the two bars is "S"

Propose an equation using which the bird will pass MINIMUM way to do this operation.
(Assume L1 is not equal to L2)

Assume that the point that the bird picks up the seed is a distance d from the first post. Now write the expression for the length of the path than the bird will follow.

Having done that you need to determine what value of d minimises this path length. If you have problems with this stage post your result for the first stage above and you will receive further assistance.

CB

3. ## Re: The bird and the seed

To CaptainBlack:

So based on what you said, we have an equation like this for the MINIMUM:
(lets assume L1 = x and L2 = y)

(y)^2 + d^2 + (S-d)^2 + (x)^2 = S
(y)^2 + d^2 + S^2 - 2Sd + d^2 + x^2 = S

(y)^2 + 2d^2 + x^2 = S - S^2 + 2Sd

and then ... ?

4. ## Re: The bird and the seed

Originally Posted by Narek
To CaptainBlack:

So based on what you said, we have an equation like this for the MINIMUM:
(lets assume L1 = x and L2 = y)

(y)^2 + d^2 + (S-d)^2 + (x)^2 = S

The left hand side is the square of the path length (which is OK we can find $d$ to minimise the square of this it will give the same answer). Why have you set this equal to $S$? If $x$ and/or $y >0$ it must be greater than $S$ .

Let $D^2$ denote the square of the path length:

$D^2=y^2 + d^2 + (S-d)^2 + x^2=d^2-2Sd+(x^2+y^2)$

This $D^2$ is what you need to find the minimum value of (and the corresponding d that gives the minimum of $D^2$).

You should have been shown a method of finding the minimum of a quadratic experssion like this.

(There is an alternative method that requires you reflect the second leg of the birds path in the ground and observe that this new path should be a straight line for the minimum length path)

CB

5. ## [SOLVED] Re: The bird and the seed

well, I didn't solve this problem and at the end, my friend told me how to solve it.

So here's how it is:

You extend the line "L1" to the bottom and draw a line from "L2" to that end (The blue line in the picture) The intersect is the MINIMUM.

Why? because if that is any other point, based on triangle inequality ( a + b > c ) the line is not minimum.

6. ## Re: [SOLVED] Re: The bird and the seed

Originally Posted by Narek
well, I didn't solve this problem and at the end, my friend told me how to solve it.

So here's how it is:

You extend the line "L1" to the bottom and draw a line from "L2" to that end (The blue line in the picture) The intersect is the MINIMUM.

Why? because if that is any other point, based on triangle inequality ( a + b > c ) the line is not minimum.

Which is the alternative method mentioned in at the end of my earlier post.

CB
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