IB Mathematics SL 5.6 Local maximum and minimum points AI SL Paper 2 - Exam Style Questions - New Syllabus

(a)
Surface Area \(A =\) Curved surface of hemisphere + Curved surface of cylinder.
\(A = \frac{1}{2}(4\pi r^2) + 2\pi rh\)
\(\mathbf{A = 2\pi r^2 + 2\pi rh}\).

(b)
Volume \(V =\) Volume of hemisphere + Volume of cylinder.
\(V = \frac{1}{2}(\frac{4}{3}\pi r^3) + \pi r^2 h\)
\(V = \frac{2\pi r^3}{3} + \pi r^2 h\)
Combine fractions: \(V = \frac{2\pi r^3 + 3\pi r^2 h}{3}\) (AG).

(c)
Given \(V = 10000\):
\(10000 = \frac{2\pi r^3 + 3\pi r^2 h}{3}\)
\(30000 = 2\pi r^3 + 3\pi r^2 h\)
\(3\pi r^2 h = 30000 – 2\pi r^3\)
\(h = \frac{30000 – 2\pi r^3}{3\pi r^2}\) (AG).

(d)
Substitute expression for \(h\) into the formula for \(A\) from part (a):
\(A = 2\pi r^2 + 2\pi r\left(\frac{30000 – 2\pi r^3}{3\pi r^2}\right)\)
\(A = 2\pi r^2 + \frac{2(30000 – 2\pi r^3)}{3r}\)
\(A = 2\pi r^2 + \frac{60000}{3r} – \frac{4\pi r^3}{3r}\)
\(A = 2\pi r^2 + \frac{20000}{r} – \frac{4\pi r^2}{3}\)
\(A = (2 – \frac{4}{3})\pi r^2 + \frac{20000}{r}\)
\(A = \frac{2\pi r^2}{3} + \frac{20000}{r}\) (AG).

(e)
Differentiate \(A\) with respect to \(r\):
\(A = \frac{2\pi}{3} r^2 + 20000 r^{-1}\)
\(\frac{dA}{dr} = \frac{4\pi r}{3} – 20000 r^{-2}\)
\(\frac{dA}{dr} = \frac{4\pi r}{3} – \frac{20000}{r^2}\).

(f)
Set \(\frac{dA}{dr} = 0\) for minimum:
\(\frac{4\pi r}{3} = \frac{20000}{r^2}\)
\(4\pi r^3 = 60000\)
\(r^3 = \frac{15000}{\pi}\)
\(r = \sqrt[3]{\frac{15000}{\pi}} \approx \mathbf{16.8}\) (16.8389…).
Substitute \(r\) back into the expression for \(h\):
\(h = \frac{30000 – 2\pi(16.8389)^3}{3\pi(16.8389)^2} \approx \mathbf{0}\).

(g)
Since \(h=0\), the cylindrical part has no height.
The best shape is a hemisphere.