急 ! F4 數 Function.

(a)
Total area of the follower bed and the jogging track
= Area of a circle with radius (10 + x) m
= π (10 + x)² m²

Hence, A(x) =π (10 + x)²


(b)
It given that the width of the jogging track exceeds 1 m.
Hence, x > 1

The diameter of the circle consisting the follower bed and the jogging trackmust NOT exceed the length of the square park.
2(10 + x) m ≤ 30 m
10 + x ≤ 15
x ≤ 5

x > 1 and x ≤ 5
Hence, domain of A(x) : 1 < x ≤ 5


(c)
When x = 2
total area of the flower bed and the jogging track
= π (10 + 2)² m²
= 144π m²

The park is a square with sides 30m long. The flower is a circle with radius 10m at the center of the square. So the closest distance from the side of the circle to the side of the square is 5m. If the jogging track is more than 5m, part of the track will be outside park. That is why x <= 5. x > 1 because the architect wants the track to be wider than 1m. So 1 < x <=5.

If you still have problem seeing this. Draw a horizontal line from the center of the circle to the side of square. The length of the line segment between the circle and the right side of the square is exactly 5. That is the track's upper limit to stay inside the park.