Question

Two posts, one 8 feet high and the other 12 feet high, stand 15 feet apart. They are to be stayed by wires attached to a single stake at ground level, the wires running to the top of the posts. Where the stake should be placed, to use the least amount of wire?

Answer 1

Answer on Question #63948 – Math – Geometry Question

Two posts, one 8 feet high and the other 12 feet high, stand 15 feet apart. They are to be stayed by wires attached to a single stake at ground level, the wires running to the top of the posts. Where the stake should be placed, to use the least amount of wire?

Solution

Let $x$ be the distance from the 12 foot pole where the wire touches the ground. The length of the wire connecting the 12 foot pole to the ground is $\sqrt{x^2 + 12^2}$ according to the Pythagorean Theorem. Since the two poles are 15 feet apart, the wire connecting the 8 foot pole will touch the ground $15 - x$ feet from the pole.

The length of this wire is $\sqrt{(15 - x)^2 + 8^2}$.

We have to minimize

$$L(x) = \sqrt{x^2 + 12^2} + \sqrt{(15 - x)^2 + 8^2}.$$

Compute the first derivative

$$L'(x) = \left(\sqrt{x^2 + 12^2} + \sqrt{(15 - x)^2 + 8^2}\right)' = \frac{x}{\sqrt{x^2 + 12^2}} - \frac{15 - x}{\sqrt{(15 - x)^2 + 8^2}}.$$

$L'(x)$ is defined for $x \in \mathbb{R}$, but we consider $0 \leq x < +15$.

$$\begin{array}{l} L'(x) = 0 \Rightarrow \frac{x}{\sqrt{x^2 + 12^2}} - \frac{15 - x}{\sqrt{(15 - x)^2 + 8^2}} = 0 \Rightarrow \\ \Rightarrow \frac{x}{\sqrt{x^2 + 12^2}} = \frac{15 - x}{\sqrt{(15 - x)^2 + 8^2}}; \\ x \sqrt{(15 - x)^2 + 8^2} = (15 - x) \sqrt{x^2 + 12^2}. \end{array}$$

In order to solve for $x$ it is easiest to square both sides and simplify:

$$\begin{array}{l} x^2 ((15 - x)^2 + 8^2) = (15 - x)^2 (x^2 + 12^2); \\ x^2 (15 - x)^2 + 64x^2 = (15 - x)^2 x^2 + 144(15 - x)^2; \end{array}$$

subtract $x^2 (15 - x)^2$ from the left-hand and the right-hand sides:

$$64x^2 = 144(15 - x)^2;$$

divide by 16 through:

$$\begin{array}{l} 4x^2 = 9(15 - x)^2; \\ 4x^2 = 9(225 - 30x + x^2); \\ 5x^2 - 270x + 2025 = 0; \end{array}$$

divide by 5:

$$\begin{array}{l} x^{2} - 54x + 405 = 0; \\ (x - 9)(x - 45) = 0, 0 \leq x \leq 15. \end{array}$$

We exclude $x = 45$. Therefore $x = 9$.

Using the First Derivative Test

If we choose $x = 0 \in [0,9)$, then

$$L'(0) = \frac{0}{\sqrt{0^{2} + 12^{2}}} - \frac{15 - 0}{\sqrt{(15 - 0)^{2} + 8^{2}}} = -\frac{15}{17} < 0.$$

If we choose $x = 15 \in (9,15]$, then

$$L'(15) = \frac{15}{\sqrt{15^{2} + 12^{2}}} - \frac{15 - 15}{\sqrt{(15 - 15)^{2} + 8^{2}}} = \frac{15}{\sqrt{369}} > 0.$$

Since the derivative changes from negative to positive, $L(x)$ will have a minimum when $x = 9$.

Therefore the stake should be placed 9 feet from the 12 foot pole.

**Answer**: the stake should be placed 9 feet from the 12 foot pole.

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