Question 11
Figure 4 shows an open container in the shape of a cylinder with radius $r \mathrm{~cm}$ and height $h \mathrm{~cm}$.
Given that the total surface area of the container is $625\pi \mathrm{~cm}^2$
(a) show that
$$ h=\frac{625-r^2}{2r} $$(3)
The volume of the container is $V \mathrm{~cm}^3$
Given that $r$ can vary,
(b) use calculus to find the value, to 3 significant figures, of $r$ for which $V$ is a maximum. Justify that this value of $r$ gives a maximum value of $V$
(6)
(c) For the value of $r$ found in part (b), find the corresponding value, to 3 significant figures, of $h$
(1)
Solution
(a) Show that $h = \frac{625 - r^2}{2r}$
The total surface area $S$ of the open cylinder is given by:
$S = \pi r^2 + 2\pi rh,$
where $\pi r^2$ is the area of the circular base and $2\pi rh$ is the area of the curved surface.
We are given that $S = 625\pi$. Substituting this:
$\pi r^2 + 2\pi rh = 625\pi.$
Divide through by $\pi$:
$r^2 + 2rh = 625.$
Rearrange to solve for $h$:
$2rh = 625 - r^2.$
Divide through by $2r$ (where $r > 0$):
$h = \frac{625 - r^2}{2r}.$
(b) Use calculus to find the value of $r$ that maximizes $V$
The volume $V$ of the cylinder is given by:
$V = \pi r^2 h.$
Substitute $h = \frac{625 - r^2}{2r}$ into the volume formula:
$V = \pi r^2 \left(\frac{625 - r^2}{2r}\right).$
Simplify:
$V = \pi \frac{r^2 (625 - r^2)}{2r}.$
$V = \frac{\pi r (625 - r^2)}{2}.$
Expand:
$V = \frac{\pi}{2} (625r - r^3).$
Differentiate $V$ with respect to $r$:
$\frac{dV}{dr} = \frac{\pi}{2} \left(625 - 3r^2\right).$
Set $\frac{dV}{dr} = 0$ to find the critical points:
$625 - 3r^2 = 0.$
Solve for $r^2$:
$3r^2 = 625.$
$r^2 = \frac{625}{3}.$
$r = \sqrt{\frac{625}{3}} = \frac{25}{\sqrt{3}} \approx 14.434.$
Thus:
$r \approx 14.4 \, \text{cm (to 3 significant figures)}.$
Justification that this value of $r$ gives a maximum
Differentiate $\frac{dV}{dr}$ again to find $\frac{d^2V}{dr^2}$:
$\frac{d^2V}{dr^2} = \frac{\pi}{2} (-6r).$
Substitute $r = 14.43$:
$\frac{d^2V}{dr^2} = \frac{\pi}{2} (-6 \cdot 14.43) = \frac{\pi}{2} (-86.4) \lt 0.$
Since $\frac{d^2V}{dr^2} \lt 0$, $V$ has a maximum at $r = 14.4$.
(c) Find the corresponding value of $h$
Using $h = \frac{625 - r^2}{2r}$, substitute $r = 14.43$:
$h = \frac{625 - (14.43)^2}{2 \cdot 14.43}.$
Simplify $14.43^2$:
$14.43^2 = 208.22.$
Substitute:
$h = \frac{625 - 208.22}{28.86}.$
$h = \frac{417.64}{28.8} \approx 14.4 \, \text{cm (to 3 significant figures)}.$
Final Answers:
- (b) $r \approx 14.4 \, \text{cm}$.
- (c) $h \approx 14.4 \, \text{cm}$.
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