IB Mathematics SL 5.6 Local maximum and minimum points AI SL Paper 2 - Exam Style Questions - New Syllabus
▶️ Answer/Explanation
Markscheme (with brief working)
(a)
Curved area of cylinder \(=2\pi xh\). Surface area of two hemispheres \(=\) surface area of a sphere \(=4\pi x^2\).
Volume condition for cylinder: \(\pi x^2 h=41 \Rightarrow h=\dfrac{41}{\pi x^2}\).
Hence \[ S=2\pi x\!\left(\frac{41}{\pi x^2}\right)+4\pi x^2 =\frac{82}{x}+4\pi x^2. \] M1 A1 A1 (AG)
Volume condition for cylinder: \(\pi x^2 h=41 \Rightarrow h=\dfrac{41}{\pi x^2}\).
Hence \[ S=2\pi x\!\left(\frac{41}{\pi x^2}\right)+4\pi x^2 =\frac{82}{x}+4\pi x^2. \] M1 A1 A1 (AG)
(b)
(i) Differentiate: \[ \frac{dS}{dx}=-\frac{82}{x^2}+8\pi x. \] A1 A1
(ii) Stationary when \(\dfrac{dS}{dx}=0\): \[ -\frac{82}{x^2}+8\pi x=0 \;\Rightarrow\;8\pi x^3=82 \;\Rightarrow\; x^3=\frac{41}{4\pi}. \] Hence \[ \boxed{a=\left(\frac{41}{4\pi}\right)^{\!1/3}}. \] M1 A1 A1
(ii) Stationary when \(\dfrac{dS}{dx}=0\): \[ -\frac{82}{x^2}+8\pi x=0 \;\Rightarrow\;8\pi x^3=82 \;\Rightarrow\; x^3=\frac{41}{4\pi}. \] Hence \[ \boxed{a=\left(\frac{41}{4\pi}\right)^{\!1/3}}. \] M1 A1 A1
(c)
(i) Second derivative: \[ \frac{d^2S}{dx^2}=\frac{164}{x^3}+8\pi. \] A1 A1
(ii) Justifying a minimum (two acceptable methods):
- Method 1 (substitution): Using \(a^3=\dfrac{41}{4\pi}\), \[ \frac{d^2S}{dx^2}\Big|_{x=a} =\frac{164}{a^3}+8\pi =164\cdot\frac{4\pi}{41}+8\pi =16\pi+8\pi =24\pi\;(>0). \] Therefore \(S\) has a local minimum at \(x=a\). M1 A1 (AG)
- Method 2 (shape of \(S”\)): Sketch or note \(\dfrac{d^2S}{dx^2}=\dfrac{164}{x^3}+8\pi>0\) for \(x>0\); thus \(S\) is convex and the stationary point at \(x=a\) is a minimum. M1 A1 (AG)
(iii) Minimum surface area \(S_{\min}=S(a)\): \[ S(a)=\frac{82}{a}+4\pi a^2 =82\!\left(\frac{4\pi}{41}\right)^{\!1/3} +4\pi\!\left(\frac{41}{4\pi}\right)^{\!2/3} =\boxed{3(4\pi)^{1/3}41^{2/3}\ \text{cm}^2} \approx \boxed{82.9\ \text{cm}^2}. \] (Exact form shown; decimal rounded to 3 s.f.) M1 A1
Total Marks: 14