Question

Two 10 meter poles are 30 meters apart. A length of wire is attached to the top of each pole and it is staked to the ground somewhere between the two poles. Where should the wire be staked so that the minimum amount of wire is used?

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94 Answers

1. Define the positions of the poles and the point where the wire is staked.

2. Let the poles be positioned at points A(0, 10) and B(30, 10) respectively.

3. Let the point where the wire is staked be at a point P(x, 0).

4. The distances from the top of each pole to the staking point P are:

PA = \sqrt{x^2 + 10^2} = \sqrt{x^2 + 100}

PB = \sqrt{(30 - x)^2 + 10^2} = \sqrt{(30 - x)^2 + 100}

6. Sum these distances to get the total length of the wire as a function of x:

L(x) = \sqrt{x^2 + 100} + \sqrt{(30 - x)^2 + 100}

7. Differentiate \(L(x)\) with respect to \(x\) and set the derivative equal to zero to find the minimum:

\frac{dL}{dx} = \frac{x}{\sqrt{x^2 + 100}} + \frac{-(30 - x)}{\sqrt{(30 - x)^2 + 100}} = 0

8. Simplify to find the critical points:

\frac{x}{\sqrt{x^2 + 100}} = \frac{30 - x}{\sqrt{(30 - x)^2 + 100}}

x \sqrt{(30 - x)^2 + 100} = (30 - x) \sqrt{x^2 + 100}

9. Square both sides to eliminate the square roots and solve for x:

x^2 (30 - x)^2 + 100x^2 = (30 - x)^2 x^2 + 100(30 - x)^2

100x^2 = 100(30 - x)^2

x^2 = (30 - x)^2

10. Solving this:

x = 30 - x

2x = 30

x = 15

Therefore, the wire should be staked x = 15 meters from either pole to minimize the total length of the wire.

2. Let the poles be positioned at points A(0, 10) and B(30, 10) respectively.

3. Let the point where the wire is staked be at a point P(x, 0).

4. The distances from the top of each pole to the staking point P are:

6. Sum these distances to get the total length of the wire as a function of x:

7. Differentiate \(L(x)\) with respect to \(x\) and set the derivative equal to zero to find the minimum:

8. Simplify to find the critical points:

9. Square both sides to eliminate the square roots and solve for x:

10. Solving this:

Therefore, the wire should be staked

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