IB Mathematics SL 5.1 Derivative interpreted as gradient function AI HL Paper 2- Exam Style Questions- New Syllabus

(a)
Total park area = \(1200\), so park width = \(\frac{1200}{x}\).
Garden length = \(x – 2(1.5) = x – 3\).
Garden width = \(\frac{1200}{x} – 2(2) = \frac{1200}{x} – 4\).
\(A = (x – 3)(\frac{1200}{x} – 4) = 1200 – 4x – \frac{3600}{x} + 12 = \mathbf{1212 – 4x – \frac{3600}{x}}\).

(b)
Set \(1212 – 4x – \frac{3600}{x} = 800 \Rightarrow 4x^2 – 412x + 3600 = 0\).
Solving the quadratic gives \(x \approx \mathbf{9.64 \text{ m}}\) (width \(\approx 124 \text{ m}\)) or \(x \approx \mathbf{93.4 \text{ m}}\) (width \(\approx 12.9 \text{ m}\)).

(c)
\(\frac{dA}{dx} = \frac{d}{dx}(1212) – \frac{d}{dx}(4x) – \frac{d}{dx}(3600x^{-1}) = \mathbf{-4 + \frac{3600}{x^2}}\).

(d)
Set \(\frac{dA}{dx} = 0 \Rightarrow \frac{3600}{x^2} = 4 \Rightarrow x^2 = 900\).
Since \(x > 3\), \(\mathbf{x = 30 \text{ m}}\).

(e)
Substitute \(x = 30\) into \(A\):
\(A = 1212 – 4(30) – \frac{3600}{30} = 1212 – 120 – 120 = \mathbf{972 \text{ m}^2}\).