“The level 2 NCEA maths question that knocked the confidence of our
top students”: <
https://www.nzherald.co.nz/nz/the-level-2-ncea-maths-question-that-knocked-the-confidence-of-our-top-students/QDISMTPN4BDD3OUELHUZ25EFOI/>.
It’s been a while since my own O-level classes (is that what NCEA is equivalent to?) But even I was able to figure out that this was an
optimization problem, solved by working out the derivative of the
surface area as a function of one linear dimension and determining the
point where that derivative was zero.
It appears that, at this level, the students are learning to integrate
and differentiate polynomials, but nothing more. For example, they
aren’t taught how to deal with negative integer exponents. That shocks
me, just a little bit.
In the following, the formulas are in Mathjax format.
----
Given volume of the containe:
$$V = \pi r^2 h$$
which is constant, minimize surface area
$$A = 2 \pi r^2 + 2 \pi r h$$
From the formula for $V$, we can determine $h$ as a function of $r$:
$$h = {V \over {\pi r^2}}$$
from which
$$A = 2 \pi {r^2 + {{2 \pi r V} \over {\pi r^2}}} = {2 \pi {r^2 + {2 V \over r}}}$$
The derivative of this is
$${{\mathrm{d}A} \over {\mathrm{d}r}} = 4 \pi r - {2 V \over {r^2}}$$
which is zero at the point of minimum area. So
$$4 \pi r - {2 V \over {r^2}} = 0$$
from which
$$2 - {V \over {\pi r^3}} = 0$$
or
$${V \over {\pi r^3}} = 2$$
We can flip that around:
$${{\pi r^3} \over V} = {1 \over 2}$$
from which
$$r^3 = {V \over {2 \pi}}$$
and we get
$$r = \sqrt[3]{V \over {2 \pi}}$$
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I did an initial derivation of the above in about 5 minutes. Then, in
writing it up, I realized I’d made two separate mistakes in the
derivation, which managed to cancel each other out. Correcting that
took another, I don’t know, 10-15 minutes.
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