Minimising the material for a closed cylinder (optimisation)

Full worked solution

1. (a) Derive the surface area formula

   First, recall the total internal surface area of a closed cylinder is the sum of the two circular end faces and the curved lateral surface:

   $$S = 2\pi r^2 + 2\pi r h$$

   We use the fixed volume constraint: the volume of a cylinder is $V=\pi r^{2}h=1600$. Rearrange to isolate $h$:

   $$h = \frac{1600}{\pi r^2}$$

   Substitute this expression for $h$ into the surface area formula:

   $$S = 2\pi r^2 + 2\pi r \cdot \frac{1600}{\pi r^2}$$

   Simplify the second term (the $\pi$ and one factor of $r$ cancel):

   $$S = 2\pi r^2 + \frac{3200}{r}$$

   This matches the required formula.

   Note.

   The $\pi r^2$ term in the volume cancels neatly when substituting into the surface area, leaving no $\pi$ in the fraction term.

2. (b) Find the stationary value of $r$

   Stationary points occur where the first derivative of $S$ with respect to $r$ equals 0. Differentiate $S$ term-by-term:

   $$\frac{dS}{dr} = \frac{d}{dr}\left(2\pi r^2\right) + \frac{d}{dr}\left(3200 r^{-1}\right) = 4\pi r - \frac{3200}{r^2}$$

   Set $\frac{dS}{dr}=0$ and solve for $r$:

   $$4\pi r - \frac{3200}{r^2} = 0$$

   Multiply through by $r^2$ to eliminate the denominator (valid for $r>0$):

   $$4\pi r^3 = 3200$$

   Rearrange to isolate $r^3$:

   $$r^3 = \frac{3200}{4\pi} = \frac{800}{\pi}$$

   Take the cube root and evaluate numerically:

   $$r = \sqrt[3]{\frac{800}{\pi}} \approx \sqrt[3]{254.65} \approx 6.3$$

   Therefore the stationary value of $r$ is 6.3 cm to 1 decimal place.

   Note.

   Always rearrange to isolate the $r^3$ term first before taking the cube root, to avoid arithmetic errors.

3. (c) Confirm the stationary point is a minimum

   We use the second derivative test to classify the stationary point. Differentiate $\frac{dS}{dr}$ again with respect to $r$:

   $$\frac{d^2 S}{dr^2} = \frac{d}{dr}\left(4\pi r - 3200 r^{-2}\right) = 4\pi + \frac{6400}{r^3}$$

   For all positive values of $r$, both terms $4\pi$ and $\frac{6400}{r^3}$ are strictly positive, so $\frac{d^{2}S}{d{r}^2}>0$ for all $r>0$. At the stationary point $r\approx 6.3$, the second derivative is positive, meaning the stationary point is a local minimum. Since this is the only stationary point for $r>0$, it is the global minimum surface area.

   Note.

   You could also use the first derivative test: check that $\frac{dS}{dr}<0$ for $r<6.3$ and $\frac{dS}{dr}>0$ for $r>6.3$, confirming a minimum.

What this tests