# A wire in the shape of Y is hanged from two points A and B, 8 feet apart. The lower end of wire reaches a point 10 feet from AB. What is the shortest possible length of the wire?

##### 2 Answers
Jun 15, 2017

Shortest length of wire that can be used is $12.31$ feet.

#### Explanation:

The above can be better described by the following figure.

Now using Pythagoras theorem, if $x$ is the length of verticle component of wire, lengths of tilted wires each is

$\sqrt{{4}^{2} + {\left(10 - x\right)}^{2}}$ or $\sqrt{16 + 100 + {x}^{2} - 20 x}$ or sqrt(x^2-20x+116

and total length of wire is $l \left(x\right) = x + 2 \sqrt{{x}^{2} - 20 x + 116}$

This will be minimum when $\frac{\mathrm{dl}}{\mathrm{dx}} = 0$

As $\frac{\mathrm{dl}}{\mathrm{dx}} = 1 + 2 \times \frac{1}{2 \sqrt{{x}^{2} - 20 x + 116}} \times \left(2 x - 20\right) = 0$

or $1 - \frac{2 \left(10 - x\right)}{\sqrt{{x}^{2} - 20 x + 116}} = 0$

or $\sqrt{{x}^{2} - 20 x + 116} = 20 - 2 x$

and squaring we get

${x}^{2} - 20 x + 116 = 400 + 4 {x}^{2} - 80 x$

or $3 {x}^{2} - 60 x + 284 = 0$

or $x = \frac{60 \pm \sqrt{3600 - 4 \times 3 \times 284}}{6}$

$= \frac{60 \pm \sqrt{3600 - 3408}}{6} = \frac{60 \pm \sqrt{192}}{6}$

or $x = 10 \pm 2.31$ i.e. $x = 12.31$ or $7.69$

But $x$ cannot be greater than $10$

Hence $x = 7.69$ and shortest length of wire that can be used is

$l \left(7.69\right) = 7.69 + 2 \sqrt{{7.69}^{2} - 20 \times 7.69 + 116}$

= $7.69 + 2 \sqrt{21.3361} = 7.69 + 4.62 = 12.31$

Jun 15, 2017

The shortest total length of the wire that can be used $= 16.92 f t$

#### Explanation:

Let the length of the perpedicular $O D$ dropped from the junction point $O$ of Y-shaped wire frame to the horizontal line AB be $x f t$.
The total height being 10ft the height of $O$ will be $\left(10 - x\right) f t$
Since in $\Delta A O B$ $O A = O B$
Here $D$ will be mid point of $A B$ So $A D = B D = 4 f t$

By pythagoras theorem

$O A = O B = \sqrt{{x}^{2} + {4}^{2}}$

So total length of the wire

$L = 2 \sqrt{{x}^{2} + {4}^{2}} + 10 - x \ldots . . \left(1\right)$

To know the minimum length of wire required we impose the condtion $\frac{\mathrm{dL}}{\mathrm{dx}} = 0$

So

$\frac{d}{\mathrm{dx}} \left(2 \sqrt{{x}^{2} + {4}^{2}} + 10 - x\right) = 0$

$\implies \left(2 \cdot \frac{1}{2} \cdot \frac{2 x}{\sqrt{{x}^{2} + {4}^{2}}} - 1\right) = 0$

$\implies \frac{2 x}{\sqrt{{x}^{2} + {4}^{2}}} = 1$

$\implies {\left(2 x\right)}^{2} = {\left(\sqrt{{x}^{2} + {4}^{2}}\right)}^{2}$

$\implies 4 {x}^{2} = {x}^{2} + 16$

$\implies x = \frac{4}{\sqrt{3}}$

Hence minimum length can be had by inserting $\implies x = \frac{4}{\sqrt{3}}$ in (1)

${L}_{\text{min}} = 2 \left(\sqrt{{\left(\frac{4}{\sqrt{3}}\right)}^{2} + {4}^{2}}\right) + 10 - \frac{4}{\sqrt{3}}$

$\implies {L}_{\text{min}} = 2 \times \frac{8}{\sqrt{3}} + 10 - \frac{4}{\sqrt{3}}$

$\implies {L}_{\text{min}} = \frac{16}{\sqrt{3}} - \frac{4}{\sqrt{3}} + 10$

$\implies {L}_{\text{min}} = \frac{12}{\sqrt{3}} + 10 = 4 \sqrt{3} + 10 = 16.92 f t$