Grazing goat problems I: sectors round a shed Problem set A Worked solutions

Two stone walls of a farmyard meet at a corner at an angle of \(\frac{\pi}{3}\). A goat is tied to the corner by a 6 m rope and grazes outside the angle between the walls. Both walls are longer than 6 m. Find the area of grass the goat can graze.

• Angle open to the animal: \(2\pi - \frac{\pi}{3} = \frac{5\pi}{3}\)
• \(A = \frac12 r^2\theta = \frac12\times 6^2\times \frac{5\pi}{3}\)
• \(A = 94.2\) m²

A goat is tied by a 5 m rope to the corner of a shed that measures 9 m by 7 m. Find the area of grass the goat can graze.

• The shed blocks a right angle, so the rope sweeps \(2\pi - \frac{\pi}{2} = \frac{3\pi}{2}\)
• \(A = \frac12\times 5^2\times\frac{3\pi}{2} = 18.75\pi\)
• \(A = 58.9\) m²

A goat is tied by a 17 m rope to a corner of a shed that measures 14 m by 9 m. Find the area of grass it can graze.

• \(\frac34\pi\times 17^2 = 216.75\pi\)
• Rope left along the 14 m wall: 17 − 14 = 3 m, giving \(\frac14\pi\times 3^2 = 2.25\pi\)
• Rope left along the 9 m wall: 17 − 9 = 8 m, giving \(\frac14\pi\times 8^2 = 16.0\pi\)
• \(A = 235\pi = 738\) m²

A shed measures 14 m by 10 m. A goat is tied by a 12 m rope to a hook on a 14 m wall, 4 m from the nearest corner. Find the area of grass the goat can graze.

• \(\frac12\pi\times 12^2 = 72\pi\)
• Left: 12 − 4 = 8 m, giving \(\frac14\pi\times 8^2 = 16\pi\)
• Right: 12 − (14 − 4) = 2 m, giving \(\frac14\pi\times 2^2 = 1\pi\)
• \(A = 89\pi = 280\) m²

A goat is tied by a 10 m rope to a corner of a shed that measures 12 m by 8 m. The rope is longer than the short wall and shorter than the long wall. Find the area of grass the goat can graze.

• \(\frac34\pi\times 10^2 = 75\pi\)
• The rope reaches the end of the 8 m wall with 10 − 8 = 2 m left. It wraps round: \(\frac14\pi\times 2^2 = 1\pi\)
• The rope is shorter than the 12 m wall, so there is no second wrap.
• \(A = 76\pi = 239\) m²

A shed is 12 m wide and 9 m deep. A goat is tied by an 11 m rope to a hook X on the front wall, x m from the left-hand corner, where \(2 \le x \le 10\).

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

   • Semicircle: \(\frac12\pi\times 11^2 = 60.5\pi\)
   • Left quarter circle: radius 11 − x. Right quarter circle: radius 11 − (12 − x) = x − 1.
   • \(A = \frac{121\pi}{2} + \frac{\pi}{4}(11-x)^2 + \frac{\pi}{4}(x-1)^2\)
   • \(= \frac{\pi}{2}(x^2 - 12x + 182)\)

2. (b) Find A when x = 3.

   • \(A = \frac{\pi}{2}(9 - 36 + 182)\)
   • \(A = 77.5\pi = 243\) m²

A goat is tied by an 11 m rope to a hook on the front wall of a shed that is 12 m wide and 9 m deep. When the hook is x m from the left-hand corner, \(2 \le x \le 10\), the grazed area is \(A = \frac{\pi}{2}(x^2 - 12x + 182)\) m². The goat grazes 230 m². Find the possible values of x.

• \(\frac{\pi}{2}(x^2 - 12x + 182) = 230\)
• Solve for x on 2 ≤ x ≤ 10: x = 5.35 or x = 6.65
• Both are in the domain. They are mirror images: 5.35 + 6.65 = 12.

Use \(A = \frac{\pi}{2}(x^2 - 12x + 182)\) for \(2 \le x \le 10\). Find where the hook should be to give the goat (a) the most grass and (b) the least grass, and the area in each case.

• Graph \(y = \frac{\pi}{2}(x^2 - 12x + 182)\) for 2 ≤ x ≤ 10.
• The vertex is at x = 6. Minimum: A = 229 m².
• The graph is a U-shape, so the maximum is at an end of the domain: x = 2 or x = 10, A = 254 m².