If the barn in Problem 45 is rectangular, $10$feet by $20$feet, what is the maximum grazing area for the cow?

The barn is rectangular by $10\times 20feet$

Consider the figure:

The maximum area = Area of circle OBDO + Area of triangle OAC+ Area of sector ACB + Area of sector OCD -Area of rectangle OEFA

The rope is $100feet,sor=100$

Area of the region role="math">OBD)=34π(100)2=23,571.429

$$\begin{gathered} \tan \angle AOF=\frac{20}{10} \\ \angle AOF={\tan }^{-1}\left(2\right) \\ =63.435° \end{gathered}$$

$$\begin{gathered} AC=100-OA \\ =100-10 \\ =90 \end{gathered}$$

$\angle AOC=\angle AOF=63.435°$

By applying Law of sines,

$$\begin{gathered} \frac{\sin \angle COA}{AC}=\frac{\sin \angle OAC}{OC}=\frac{\sin \angle ACO}{OA} \\ \frac{\sin 63.435°}{90}=\frac{\sin \angle OAC}{OC}=\frac{\sin \angle ACO}{10} \end{gathered}$$

Combine first and last,

$$\begin{gathered} \sin \angle ACO=\frac{63.435°}{9} \\ \angle ACO={\sin }^{-1}(0.0993808) \\ \angle ACO=5.703° \end{gathered}$$

By Interior angle property of triangles,

$$\begin{gathered} \angle AOC+\angle OAC+\angle OCA=180° \\ \angle OAC=180°-63.435°-5.703° \\ \angle OAC=110.862° \end{gathered}$$

Area of $∆OAC=\frac{1}{2}\left(OA\right)\left(AC\right)\sin \angle OAC$
$$\begin{gathered} =\frac{1}{2}\left(10\right)\left(90\right)\sin 110.862° \\ =420.4984 \end{gathered}$$

$$\begin{gathered} \angle OAC+\angle CAB=180° \\ \angle CAB=180°-110.862° \\ =69.138° \end{gathered}$$

Area of the sector $ACB=\frac{angle}{360°}\times {\mathrm{\pi r}}^2$
localid="1647440938685">=69.138°360°×π(90)2=4889.044

$$\begin{gathered} \angle COD=90°-\angle AOF \\ =90°-63.435° \\ =26.565° \end{gathered}$$

Area of the sector $=\frac{26.565°}{360°}\times \mathrm{\pi }{\left(100\right)}^2$
$=2319.1667$

Area of rectangle$=20\times 10$

$=200$

Area grazed by the cow$=23571.429+420.4984+4889.044+2319.667-200$

$=31,000.14$