A horse is tied to an outside corner of a 20 foot by 10 foot rectangular barn. What is the maximum area the horse can graze outside, if the rope has length *L*? Solve for *L* = 5 feet and *L* = 25 feet, and find the exact answer. If you are up for a challenge, solve for *L* = 50 feet. You can solve for the answer to 3 decimal places.

Can you figure it out? Watch the video for a solution.

**Can You Solve The Horse Grazing Puzzle?**

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.**Answer To The Horse Grazing Puzzle**

If the rope has *L* = 5 feet, then the horse can graze the area of 3/4 of a circle with radius 5.

The horse can graze an area (3/4)π(5^{2}) = 75π/4 ≈ 58.905 square feet.

If the rope has *L* = 25 feet, then the horse can graze the area of 3/4 of a circle with radius 25. But along the 10 foot wall, there is also 15 feet of rope left over. The horse can then graze an extra 1/4 circle with a radius of 15 feet. Similarly, along the 20 foot wall, the horse can also graze an extra 1/4 circle with a radius of 5 feet.

The horse can graze an area equal to the sum of these shapes. The horse can therefore graze an area (1/4)π[3(25^{2}) + (15^{2}) + (5^{2})] = 2125π/4 ≈ 1668.971 square feet.

The case of *L* = 50 feet is more complicated. The horse can graze 3/4 of a circle with a radius of 50. Along the 10 foot wall, the rope has an extra 40 feet, so the horse can reach an extra 1/4 circle with a radius of 40. And along the 20 foot wall, the rope has an extra 30 feet, so the horse can reach an extra 1/4 circle with a radius of 30. But the extra 1/4 circles overlap which makes the problem much harder to solve.

I will present two ways to solve for the area. One method is to use calculus to solve for the overlapping area directly. The other method is to use trigonometry to dissect the area into non-overlapping shapes.

**Calculus approach to L = 50**

If we add up the areas of the 1/4 circles, we would be double-counting the area of the region where the circles overlap. If we can solve for the area of the overlap, we could subtract that out.

grazing area = (area of 3/4 circle radius 50) + (area of 1/4 circle radius 40) + (area of 1/4 circle radius 30) – (overlap of 1/4 circles)

What is the area of the overlapping regions?

We can put a coordinate system and then set up integrals. The origin is placed at the left corner where the barn intersects the circle with a radius of 40. We then have one circle with a radius of 40 centered at the origin, and we have another circle with a radius of 30 centered at (20, -10). We can write equations for the *y* values of these circles (we only care about the semi-circles above the centers of the circles, so we can disregard the negative square roots):

*x*^{2} + *y*^{2} = 40^{2}*y* = √(1600 – *x*^{2})

(*x* – 20)^{2} + (*y* + 10)^{2} = 30^{2}*y* = -10 + √(900 – (*x* – 20)^{2})

We can set the equations for these curves equal to each other, and we can solve that they intersect at *x* = 24 + 4√11.

We now focus on only the overlapping region to illustrate the overlap is the sum of two definite integrals.

We have solved the overlapping region is approximately 341.77937.

We can now go back and solve for the total grazing area:

grazing area = (area of 3/4 circle radius 50) + (area of 1/4 circle radius 40) + (area of 1/4 circle radius 30) – (overlap of 1/4 circles)

grazing area = (π/4)[3(50)^{2} + 40^{2} + 30^{2}] – (341.77937)

grazing area = 2500π – (341.77937) ≈ 7512.202

**Trigonometry approach to L = 50**

The problem can also be solved without calculus, although there are more steps. The key insight is to connect the point of intersection between the circles to opposite corners of the barn.

This creates a kite-like shaped quadrilateral. Above the kite-like shaped quadrilateral is a circular sector from the circle with radius 40. And below the quadrilateral is another circular sector from the circle with radius 30. We can solve for the angles of these circular sectors and then solve for their areas.

In other words, we can solve for the total area by adding up the areas of 3 non-overlapping shapes.

First we solve for the quadrilateral area. Let’s add 1/2 of the rectangular barn to make a large triangle. The diagonal of the rectangular barn is 10√5 because it is the hypotenuse of a right triangle with legs 10 and 20. The large triangle has sides of 40 and 30 corresponding to the radii of the circles. The area of the large triangle can then be found using Heron’s formula. And the area of 1/2 the barn is also readily calculated.

We have found the quadrilateral has an area of 100√11 – 100.

Now we need to find the angles for the circular sectors. We can use trigonometry.

We can find the angles of 1/2 of the barn readily using the inverse cosine of an adjacent leg over the hypotenuse. Then we calculate the angles of the large triangle using the law of cosines.

The angle of a circular sector with 40 is found by subtracting the upper angle from the right triangle and the angle from the large triangle opposite the side equal to 30. The angle of a circular sector with 30 is found by subtracting the lower right angle from the right triangle and the angle from the large triangle opposite the side equal to 40.

We can use those angles to calculate the area of the circular sectors, which then leads to the total area of the region.

This region has an area of approximately 1621.716. We add that to the 3/4 circle with a radius of 50 to get the total grazing area.

grazing area = (3/4)π(50)^{2} + 1621.716 ≈ 7512.202

We again get to the same answer without using calculus.

**Sources**

Math StackExchange

http://math.stackexchange.com/questions/1942222/a-goat-tied-to-a-corner-of-a-rectangle

Sketch of trigonometry approach

http://mathcentral.uregina.ca/QQ/database/QQ.09.08/h/misty1.html

Another sketch of the trigonometry approach

http://mathforum.org/dr.math/faq/faq.grazing.html

See the original post:

Can You Solve The Horse Grazing Area? Sunday Puzzle