A calculus of the absurd

Chapter C Assorted problems

Solutions to problems which have yet to be categorised.

C.0.1 Cones problem

Problem: Two similar cones, \(X\) and \(Y\), have surface areas \(270\text {cm}^2\) and \(120\text {cm}^2\) respectively.

The volume of cone \(X\) is \(1215\text {cm}^3\).

Show that the volume of cone Y is \(360 \text {cm}^3\).


We have two cones, let’s let \(X\) be this one, with radius \(r\) and height \(h\)

(-tikz- diagram)

Then we can draw \(Y\) (which is smaller than \(X\), but not drawn to scale here). As they are similar cones, the radius of \(Y\) is equal to \(ar\) (where \(a\) is the scale factor between the two cones) and the height of \(Y\) is equal to \(ah\)

(-tikz- diagram)

First, the volume of a cone is \(\frac {\pi r^2 h}{3}\) and the surface area of a cone is \(\pi r (r + \sqrt {h^2 + r^2})\). These can be derived using integration (but not in Single Maths A Level at least).

Using this, we can write this equation relating the radius and height of \(X\) to the value given in the question

\begin{equation} \pi r(r + \sqrt {h^2 + r^2}) = 270 \end{equation}

We can also do the same for \(Y\)

\begin{equation} \pi a r(ar + \sqrt []{(ah)^2+(ar)^2}) = 120 \end{equation}

This can be simplified a bit by factoring the \(a^2\) from the square rooted part of the equation, giving

\begin{equation} \pi a r(ar + \sqrt {a^2}\sqrt []{(h)^2+(r)^2}) = 120 \end{equation}

and thus that

\begin{equation} \pi a^2 r(r + \sqrt {(h)^2+(r)^2}) = 120 \end{equation}

We can solve this for \(a\) by dividing the two equations, giving that

\begin{equation} \frac { \pi a^2 r(r + \sqrt {(h)^2+(r)^2}) }{\pi r(r + \sqrt {h^2 + r^2})} = \frac {120}{270} \end{equation}

and thus, after cancelling, that

\begin{equation} a^2 = \frac {9}{4} \end{equation}

which means that

\begin{equation} a = \frac {3}{2} \end{equation}

Given this, we can then turn to the volume formula.

We know that the volume of \(X\) is

\begin{equation} \label {volume of x} \pi r^2 h = 1215 \end{equation}

and the formula for the volume of \(Y\) is equal to

\begin{equation} \pi (ar)^2 (ah) = a^3 \pi r^2 h \end{equation}

Thus, by multiplying Equation C.8 through by \(a^3\), we have

\begin{equation} a^3 \pi r^2 h = a^3 1215 \end{equation}

The left-hand side is just the volume of \(Y\), and the right-hand side is just

\begin{align} a^3 1215 &= \left (\frac {2}{3}\right )^3 \cdot 1215 \\ &= 360 \end{align}

Which is what we needed to show.