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2009 Paper 1 Q5
D: 1484.0 B: 1484.0
optimisation cone implicit differentiation constrained optimisation stationary points Pythagorean relation volume surface area

A right circular cone has base radius \(r\), height \(h\) and slant length \(\ell\). Its volume \(V\), and the area \(A\) of its curved surface, are given by \[ V= \tfrac13 \pi r^2 h \,, \ \ \ \ \ \ \ A = \pi r\ell\,. \]

  1. Given that \(A\) is fixed and \(r\) is chosen so that \(V\) is at its stationary value, show that \(A^2 = 3\pi^2r^4\) and that \(\ell =\sqrt3\,r\).
  2. Given, instead, that \(V\) is fixed and \(r\) is chosen so that \(A\) is at its stationary value, find \(h\) in terms of \(r\).

Solution:

  1. Given \(A\) is fixed, and \(h^2 + r^2 = \ell^2\), we can look at \begin{align*} && V^2 &= \frac19 \pi^2 r^4 h^2 \\ &&&= \frac19\pi^2r^4(\ell^2 - r^2) \\ &&&= \frac19\pi^2 r^4\left (\frac{A^2}{\pi^2r^2} - r^2 \right) \\ &&&= \frac{A^2r^2 - \pi^2r^6}{9} \end{align*} Differentiating wrt to \(r\) we find that \(2rA^2-6\pi^2 r^5 = 0\) or hence \(A^2 = 3\pi^2 r^4 \Rightarrow A = \sqrt{3}\pi r^2\). Therefore \(\sqrt{3}\pi r^2 = \pi r \ell \Rightarrow \ell = \sqrt{3}r\).
  2. Supposing \(V\) is fixed, then \begin{align*} && A^2 &= \pi^2 r^2\ell^2 \\ &&&= \pi^2 r^2 (h^2+r^2) \\ &&&= \pi^2 r^2 \left ( \frac{9V^2}{\pi^2r^4} + r^2 \right) \\ &&&= 9V^2r^{-2} + \pi^2r^4 \\ \end{align*} Differentiating wrt to \(r\) we find \(-18V^2r^{-3} + 4\pi^2 r^3 = 0\) so \(V^2 = \frac{2\pi^2}{9}r^6\) or \(V = \frac{\sqrt{2}\pi}{3}r^3\), from which it follows: \(\frac{\sqrt{2}\pi}{3}r^3 = \frac13\pi r^2 h \Rightarrow h = \sqrt{2}r\)

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2001 Paper 2 Q3
D: 1600.0 B: 1500.0
vectors cuboid scalar product angle between diagonals optimisation inequality surface area volume geometric proof

The cuboid \(ABCDEFGH\) is such \(AE\), \(BF\), \(CG\), \(DH\) are perpendicular to the opposite faces \(ABCD\) and \(EFGH\), and \(AB =2, BC=1, AE={\lambda}\). Show that if \(\alpha\) is the acute angle between the diagonals \(AG\) and \(BH\) then $$\cos {\alpha} = |\frac {3-{\lambda}^2} {5+{\lambda}^2} |$$ Let \(R\) be the ratio of the volume of the cuboid to its surface area. Show that \(R<\frac{1}{3}\) for all possible values of \(\lambda\). Prove that, if \(R\ge \frac{1}{4}\), then \(\alpha \le \arccos \frac{1}{9}\).

Solution:

TikZ diagram
Set \(A\) to be the origin, then \(B = \langle 2, 0, 0 \rangle, G = \langle 2, 1, \lambda \rangle, H = \langle 0, 1, \lambda \rangle\), in particular \begin{align*} && AG&= \langle 2, 1, \lambda \rangle \\ && BH &= \langle -2, 1, \lambda \rangle \\ \Rightarrow && \cos \alpha &= |\frac{-4+1+\lambda^2}{\sqrt{2^2+1^2+\lambda^2}\sqrt{(-2)^2+1^2+\lambda^2}}| \\ &&&= |\frac{-3+\lambda^2}{5+\lambda^2}| \end{align*} \begin{align*} && \text{Volume} &= 2\lambda \\ && \text{Surface area} &= 2\cdot2\lambda + 2\cdot\lambda + 2\cdot2 \\ \Rightarrow && R&= \frac{\lambda}{3\lambda + 2} < \frac{1}{3} \\ && \frac14 &\leq R \\ \Rightarrow && 3\lambda +2 &\leq 4\lambda \\ \Rightarrow &&2 & \leq \lambda \end{align*} Then \(\frac{\lambda^2-3}{5+\lambda^2}\) is increasing as \(\lambda\) increases, in particularly the smallest value is \(\frac{1}{9}\).

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1996 Paper 1 Q1
D: 1484.0 B: 1500.0
optimisation differentiation cylinder surface area volume sphere geometric proof AM-GM inequality

A cylindrical biscuit tin has volume \(V\) and surface area \(S\) (including the ends). Show that the minimum possible surface area for a given value of \(V\) is \(S=3(2\pi V^{2})^{1/3}.\) For this value of \(S\) show that the volume of the largest sphere which can fit inside the tin is \(\frac{2}{3}V\), and find the volume of the smallest sphere into which the tin fits.

Solution: Suppose we have height \(h\) and radius \(r\), then: \(V = \pi r^2 h\) and \(S = 2\pi r^2 + 2\pi r h\). \(h = \frac{V}{\pi r^2}\), so \begin{align*} S &= 2 \pi r^2 + 2 \pi r\frac{V}{\pi r^2} \\ &= 2\pi r^2 +V \frac1{r}+V \frac1{r} \\ &\underbrace{ \geq }_{\text{AM-GM}} 3 \sqrt[3]{2\pi r^2 \frac{V^2}{r^2} } = 3 (2 \pi V^2)^{1/3} \end{align*} Equality holds when \(r = \sqrt[3]{\frac{V}{2 \pi}}, h = \frac{V}{\pi (V/2\pi)^{2/3}} = \sqrt[3]{\frac{4V}{\pi}}\) Since \(h > r\) the sphere has a maximum radius of \(r\) and so it's largest volume is \(\frac43 \pi r^3 = \frac43 \pi \frac{V}{2 \pi} = \frac23 V\).

TikZ diagram
The radius of the sphere is \(\sqrt{\left (\frac{r}{2} \right)^2 + \left (\frac{h}{2} \right)^2 } = \frac12 \sqrt{r^2+h^2}\) \begin{align*} V_{sphere} &= \frac43 \pi (r^2+h^2)^{3/2} \\ &= \frac43 \pi \left (\left( \frac{V}{2 \pi} \right)^{2/3}+\left( \frac{4V}{ \pi} \right)^{2/3} \right)^{3/2} \\ &= \frac43 \pi \frac{V}{ \pi} \left ( 2^{-2/3}+4^{2/3}\right)^{3/2} \\ &= \frac 43 V \left ( \frac{1+4}{2^{2/3}} \right)^{3/2} \\ &= \frac43 \frac{5^{3/2}}{2} V \\ &= \frac{2 \cdot \sqrt{125}}{3} V \end{align*}

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1994 Paper 1 Q1
D: 1500.0 B: 1500.0
coordinate geometry 3D geometry volume surface area trigonometry optimisation roof ridge practical modelling

My house has an attic consisting of a horizontal rectangular base of length \(2q\) and breadth \(2p\) (where \(p < q\)) and four plane roof sections each at angle \(\theta\) to the horizontal. Show that the length of the roof ridge is independent of \(\theta\) and find the volume of the attic and the surface area of the roof.

Solution:

TikZ diagram
The distance to the top of the house (viewed from above) from the long side will be \(p\). The distance from the short side will also be the same, since the roof sections are climbing at the same angle, so they will take just as far to reach the top. Therefore the length of the roof ridge will be \(2q - 2p\) which is independent of \(\theta\). \vspace{1em} The height of the roof will be \(h = p \tan \theta\). The attic can be split into a prism (along the roof ridge) and a pyramid (along the sloping sides). The pyramid will have volume \(\frac13 p \tan\theta (2p)^2 = \frac83 \tan\theta p^3\). The prism will have volume \(2(q-p)p^2 \tan\theta\). Therefore the total volume will be \(\l \frac{2}{3}p + 2q \r p^2\tan\theta \) The distance (along the plane) to the roof of the house will be \(\frac{p}{\cos \theta}\) and therefore the two end roof-sections will be triangles of area \(\frac{p^2}{\cos \theta}\). The two side roof-sections will be trapiziums will area \(\frac{1}{2} \l 2q + 2(q-p) \r \frac{p}{\cos \theta}\) Therefore the total area will be \(\frac{1}{\cos \theta} \l 2p^2 + 4pq - 2p^2 \r = \frac{4pq}{\cos \theta}\)

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1991 Paper 1 Q2
D: 1516.0 B: 1469.6
differentiation related rates sphere volume surface area optimisation proportionality melting

Frosty the snowman is made from two uniform spherical snowballs, of initial radii \(2R\) and \(3R.\) The smaller (which is his head) stands on top of the larger. As each snowball melts, its volume decreases at a rate which is directly proportional to its surface area, the constant of proportionality being the same for both snowballs. During melting each snowball remains spherical and uniform. When Frosty is half his initial height, find the ratio of his volume to his initial volume. If \(V\) and \(S\) denote his total volume and surface area respectively, find the maximum value of \(\dfrac{\mathrm{d}V}{\mathrm{d}S}\) up to the moment when his head disappears.

Solution: \(V_h = \frac43 \pi r_h^3, S_h = 4 \pi r_h^2\) \(\frac{\d V_h}{\d t} = -k4\pi r_h^2 \Rightarrow 4\pi r_h^2 \frac{\d r_h}{\d t} = -k 4\pi r_h^2 \Rightarrow \frac{\d r_h}{\d t} = -k\) Therefore \(r_h = 2R - kt, r_b = 3R - kt\). The height will halve when \(2kt = \frac{5}{2}R \Rightarrow kt = \frac{5}{4}R\) and the two sections will have radii \(\frac{3}{4}R\) and \(\frac{7}{4}R\) and the ratio of the volumes will be: \begin{align*} \frac{\frac{3^3}{4^3}+\frac{7^3}{4^3}}{2^3+3^3} = \frac{37}{224} \end{align*} \begin{align*} && \frac{\d V}{\d t} &= -4\pi k(r_h^2+r_b^2) \\ && \frac{\d S}{\d t} &= -8\pi k (r_h+r_b) \\ \Rightarrow && \frac{\d V}{\d S} &= \frac{r_h^2 + r_b^2}{2(r_h+r_b)} \\ &&&= \frac{(2R-kt)^2+(3R-kt)^2}{2(5R-2kt)} \\ &&&= \frac{13R^2-10Rkt+2k^2t^2}{2(5R-2kt)} \\ &&&= \frac{13R^2-10Rs + 2s^2}{2(5R-2s)} \end{align*} Where \(s = kt\) and \(0 \leq s \leq 2R\). We can maximise this but differentiating wrt to \(s\). \begin{align*} \Rightarrow && &= \frac{(-10R+4s)(10R-4s)+4(13R^2-10Rs+2s^2)}{4(5R-2s)^2} \\ &&&= \frac{-48R^2+40Rs-8s^2}{4(5R-2s)^2} \\ &&&= \frac{-8(s-2R)(s-3R)}{4(5R-2s)^2} \\ &&&<0 \end{align*} Therefore it is largest when \(s = 0\), ie \(\frac{13R^2}{10R} = \frac{13}{10}R\)

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1989 Paper 3 Q1
D: 1700.0 B: 1516.0
coordinate geometry surface area sphere zone of sphere circle tangent line surface of revolution cone arc length

Prove that the area of the zone of the surface of a sphere between two parallel planes cutting the sphere is given by \[ 2\pi\times(\mbox{radius of sphere})\times(\mbox{perpendicular distance between the planes}). \] A tangent from the origin \(O\) to the curve with cartesian equation \[ (x-c)^{2}+y^{2}=a^{2}, \] where \(a\) and \(c\) are positive constants with \(c>a,\) touches the curve at \(P\). The \(x\)-axis cuts the curve at \(Q\) and \(R\), the points lying in the order \(OQR\) on the axis. The line \(OP\) and the arc \(PR\) are rotated through \(2\pi\) radians about the line \(OQR\) to form a surface. Find the area of this surface.

Solution:

TikZ diagram
We can choose a coordinate frame where the parallel planes are parallel to the \(y-z\) axis. Then we can compute the surface area as an integral of the surface of revolution for \(x^2 + y^2 = r^2\). Using \(y = r \sin t, x = r \cos t\) we have: \begin{align*} S &= 2\pi\int_{\cos^{-1}a}^{\cos^{-1}b}y \sqrt{\left ( \frac{\d x}{\d t} \right)^2+\left ( \frac{\d y}{\d t} \right)^2} \d t \\ &=2\pi\int_{\cos^{-1}a}^{\cos^{-1}b} r^2 \sin t \d t \\ &= 2\pi \cdot r^2 \cdot (a - b) \\ &= 2 \pi \cdot r \cdot (ra-rb) \\ &= 2\pi\times(\mbox{radius of sphere})\times(\mbox{perpendicular distance between the planes}). \end{align*}
TikZ diagram
We can view this surface as a sphere missing a cap of height \(XQ\) and adding a cone of slant height \(OP\) and radius \(PX\) The centre of the circle is at \((c,0)\) and \(OP^2 + a^2 = c^2 \Rightarrow OP = \sqrt{c^2-a^2}\) Since \(OPC \sim OXP\) we must have that \(\frac{OX}{OP} = \frac{OP}{OC} \Rightarrow OX = \frac{c^2-a^2}{c}\) and \(\frac{PX}{OP} = \frac{CP}{OC} \Rightarrow PX = \frac{a}{c}\sqrt{c^2-a^2}\) \(QX = OX - OQ = \frac{c^2-a^2}{c}-(c-a) = \frac{ac-a^2}{c}\) Therefore the surface area is: \begin{align*} S &= 4 \pi a^2 - 2\pi \cdot a \cdot QX+ \pi PX \cdot OP \\ &= 4 \pi a^2 - 2\pi a \cdot \frac{ac-a^2}{c}+\pi \frac{a}{c}\sqrt{c^2-a^2}\cdot \sqrt{c^2-a^2} \\ &= 4\pi a^2 -2\pi \frac{a^2c-a^3}{c}+\pi \frac{ac^2-a^3}{c} \\ &= \pi a \frac{(a+c)^2}{c} \end{align*}

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