Grazing goat problems I: sectors round a shed Problem set B One question at a time

Two walls of a barn yard meet at a corner at an angle of 1.2 radians. A cow is tied to the corner by a 7.5 m rope and grazes outside the angle between the walls. Both walls are longer than 7.5 m. Find the area of grass the cow can graze.

• Angle open to the animal: \(2\pi - 1.2 = 5.083\)
• \(A = \frac12 r^2\theta = \frac12\times 7.5^2\times 5.083\)
• \(A = 143\) m²

A llama is tied by a 6 m rope to the corner of a greenhouse that measures 10 m by 8 m. Find the area of grass the llama can graze.

• The shed blocks a right angle, so the rope sweeps \(2\pi - \frac{\pi}{2} = \frac{3\pi}{2}\)
• \(A = \frac12\times 6^2\times\frac{3\pi}{2} = 27.0\pi\)
• \(A = 84.8\) m²

A donkey is tied by a 13 m rope to a corner of a barn that measures 11 m by 7 m. Find the area of grass it can graze.

• \(\frac34\pi\times 13^2 = 126.75\pi\)
• Rope left along the 11 m wall: 13 − 11 = 2 m, giving \(\frac14\pi\times 2^2 = 1.0\pi\)
• Rope left along the 7 m wall: 13 − 7 = 6 m, giving \(\frac14\pi\times 6^2 = 9.0\pi\)
• \(A = 136.75\pi = 430\) m²

A summer house measures 15 m by 8 m. A ram is tied by an 11 m rope to a hook on a 15 m wall, 5 m from the nearest corner. Find the area of grass the ram can graze.

• \(\frac12\pi\times 11^2 = 60.5\pi\)
• Left: 11 − 5 = 6 m, giving \(\frac14\pi\times 6^2 = 9\pi\)
• Right: 11 − (15 − 5) = 1 m, giving \(\frac14\pi\times 1^2 = 0.25\pi\)
• \(A = 69.75\pi = 219\) m²

A pony is tied by a 12 m rope to a corner of a stable that measures 15 m by 9 m. The rope is longer than the short wall and shorter than the long wall. Find the area of grass the pony can graze.

• \(\frac34\pi\times 12^2 = 108\pi\)
• The rope reaches the end of the 9 m wall with 12 − 9 = 3 m left. It wraps round: \(\frac14\pi\times 3^2 = 2.25\pi\)
• The rope is shorter than the 15 m wall, so there is no second wrap.
• \(A = 110.25\pi = 346\) m²

A barn is 14 m wide and 10 m deep. A donkey is tied by a 12 m rope to a hook X on the front wall, x m from the left-hand corner, where \(2 \le x \le 12\).

1. (a) Find an expression for the grazed area A in terms of x.

   • Semicircle: \(\frac12\pi\times 12^2 = 72\pi\)
   • Left quarter circle: radius 12 − x. Right quarter circle: radius 12 − (14 − x) = x − 2.
   • \(A = \frac{144\pi}{2} + \frac{\pi}{4}(12-x)^2 + \frac{\pi}{4}(x-2)^2\)
   • \(= \frac{\pi}{2}(x^2 - 14x + 218)\)

2. (b) Find A when x = 5.

   • \(A = \frac{\pi}{2}(25 - 70 + 218)\)
   • \(A = 86.5\pi = 272\) m²

A donkey is tied by a 12 m rope to a hook on the front wall of a barn that is 14 m wide and 10 m deep. When the hook is x m from the left-hand corner, \(2 \le x \le 12\), the grazed area is \(A = \frac{\pi}{2}(x^2 - 14x + 218)\) m². The donkey grazes 280 m². Find the possible values of x.

• \(\frac{\pi}{2}(x^2 - 14x + 218) = 280\)
• Solve for x on 2 ≤ x ≤ 12: x = 3.96 or x = 10.0
• Both are in the domain. They are mirror images: 3.96 + 10.0 = 14.

Use \(A = \frac{\pi}{2}(x^2 - 14x + 218)\) for \(2 \le x \le 12\). Find where the hook should be to give the donkey (a) the most grass and (b) the least grass, and the area in each case.

• Graph \(y = \frac{\pi}{2}(x^2 - 14x + 218)\) for 2 ≤ x ≤ 12.
• The vertex is at x = 7. Minimum: A = 265 m².
• The graph is a U-shape, so the maximum is at an end of the domain: x = 2 or x = 12, A = 305 m².