A goat is tied to a rectangular shed on the side of the shed. The rope is 3 metres long. How much grass can the goat eat?

Question

Shed is 3m x 4m ( rectangle ) and the chain is 3m long. Goat cannot be tied to the corner of the shed.

Answer 1

the goat can reach a semi-circle of radius 3 on the side on which it is tied.

 

.5π(3)² = 4.5π

 

around each of the two corners, the goat can reach another quarter circle of radius 3 - x,

where x is the distance from the corner to the tie off point. If he is tied in the middle of the 4' side,

then x = 2 and the quarter circles are equal and make a second semi circle of radius 1.

 

.5π(1)² = .5π

 

If he is tied in the middle of the 3' side, then x = 1.5.

 

.5π(1.5)² = 1.125π

 

So you can see, the exact location of the tie off point does make a difference. "Not a corner" is not

enough information.

Answer 2

If the rope was attached to the middle of the shed on the long side then:

 

A1 = π*(R^{2/2;+ r2/2)   where: A1 = total area of ground with grass density assumed the same on ground anywhere}

                                                  so the grass the goat eats is measured in ground area (m²)

                                          R = length of rope = 3 m

                                          r  = 1 m of rope (extends past corner of shed on both sides)

 

A1 = π*(4.5 m² +  0.5 m²) =  π*5 m²

 

If the rope was attached to the middle of the shed on the short side then:

A2 = π*(R²²/2)        where: A2 = total area of ground 
                                                 R = length of rope = 3 m
                                                 r2 = 1.5 m of rope (extends past corner of shed on both sides)

 

A2 = π*(4.5 m²²/2)  =  π*(4.5 m² + 1.125 m²)  = π*5.625 m² 

 

A2 > A1

 

If the rope was attached to 1 m from the shed corner on the short side then:

A3 = π*(R^{2/2 + r32/4 + r42/4)    where: A2 = total area of ground 
                                                         R = length of rope = 3 m
                                                         r3 = 2 m of rope (extends past corner of shed on one side)}

                                                         r4 = 1 m of rope (extends past corner of shed on the other side)

A3 = π*(4.5 m² + 4 m²/4  + 1 m²/4) = π*(4.5 m² + 1 m² + 0.25 m²) = π*5.75 m²

 

 

A3 > A2 > A1

 

 If the rope was attached to 0.5 m from the shed corner on the short side then:

A4 = π*(R²²/4 + r6²/4)     where: A4 = total area of ground                               &,nbsp;                           R = length of rope = 3 m
                                                          r5 = 2.5 m of rope (extends past corner of shed on one side)
                                                          r6 = 0.5 m of rope (extends past corner of shed on the other side)

A4 = π*(4.5 m² + 6.25 m²/4 + 0.25 m²/4) = π*(4.5 m² + 1.5625 m² + 0.0625 m²) = π*6.125 m²

A4 > A3 > A2 > A1

 

Therefore, if the rope is tied closer to the shed corner on the short side of the shed then the goat would get the most grass. 

 

A4 =  π*6.125 m² = 19.24 m² of grass area the goat can eat