# A goat grazes a rectangular paddock 10 metres by 20 metres. It is tethered to the fence at one corner of the paddock by an inextensible rope of length x metres, where 10<x<20? (More in 'details' section).

## A goat grazes a rectangular paddock 10 metres by 20 metres. It is tethered to the fence at one corner of the paddock by an inextensible rope of length $x$ metres, where $10 < x < 20$. (a) Show that the goat can graze an area of $A$ ${m}^{2}$, where $A = \frac{1}{2} {x}^{2} {\sin}^{- 1} \left(\frac{10}{x}\right) + 5 \sqrt{{x}^{2} - 100}$ (b) If the goat can graze an area equal to half the area of the paddock, find the length of the rope to one decimal place using one application of Newton's method. Thanks!

Oct 5, 2017

Let the rope of length $x$ m subtends angles $\theta$ radian with the lengths $A D \mathmr{and} B C$ of the rectangular paddock ABCD in its particular position as shown in the above figure.

It is given that $10 m < x < 20 m$

The maximum circular area that the goat can access when it is tethered by an in-extensible rope of length x metre is $= \pi {x}^{2}$ ${m}^{2}$

(a) The total possible area $A$ of grazing within the rectangular paddock will be

$A = \text{area of sector} A P Q + \Delta A Q B$

$\implies A = \frac{\pi {x}^{2}}{2 \pi} \times \theta + \frac{1}{2} \times B P \times A B$

$\implies A = \frac{1}{2} {x}^{2} \times {\sin}^{-} 1 \left(\frac{10}{x}\right) + \frac{1}{2} \times \sqrt{{x}^{2} - 100} \times 10 {m}^{2}$

$\implies A = \frac{1}{2} {x}^{2} {\sin}^{-} 1 \left(\frac{10}{x}\right) + 5 \sqrt{{x}^{2} - 100} {m}^{2}$

(b) If the goat can graze an area equal to half the area of the paddock $\left(100 {m}^{2}\right)$, then we get the following equation putting $A = 100 {m}^{2}$ .

$\implies 100 = \frac{1}{2} {x}^{2} {\sin}^{-} 1 \left(\frac{10}{x}\right) + 5 \sqrt{{x}^{2} - 100} {m}^{2}$

$f \left(x\right) = \frac{1}{2} {x}^{2} {\sin}^{-} 1 \left(\frac{10}{x}\right) + 5 \sqrt{{x}^{2} - 100} - 100 = 0$

The length of the rope to one decimal place calculated using one application of Newton's method will be $x \approx 11.7$m.

This calculation can be done by online calculator .(Link given below)

Taking

$f \left(x\right) = \frac{1}{2} {x}^{2} {\sin}^{-} 1 \left(\frac{10}{x}\right) + 5 \sqrt{{x}^{2} - 100} - 100$

and

$f ' \left(x\right) = x {\sin}^{-} 1 \left(\frac{10}{x}\right) - \frac{5}{\sqrt{1 - \frac{100}{x} ^ 2}} + \frac{5 x}{\sqrt{{x}^{2} - 100}}$