While doing a newspaper problem section yesterday, I came upon the following question.
A box has a length four times its height, which is a third of its width. If its volume is 768 cubic centimetres, what is the width of the box?
Naturally, I set about it as follows:
$$lhw=768$$ $$l=4h=\frac{w}{3}$$
From this follows:
$$3l=12h=w$$ $$h*4h*12h=768$$
Now, $$12h*4h*h=48h^2*h=48h^3=768$$
From this, $$\frac{768}{48}=h^3=16$$ $$\sqrt[3]{16}=2.51984209979=h$$
If Inserting this into the problem yields: $$(h)2.51984209979*(4h)10.07936839916*(12h)30.23810519748=768.00000000024$$
So, it seems as if my solution is nearly optimal, with
$$w=12h=30.23810519748$$
Yet, the solution given in the newspaper is $w=12$. From a naive perspective, this seems ludicrous.
If $$w=12$$ $$3l=12h=w$$ Then $$l=4, h=1$$ But $$12*4*1\neq768$$
Where is my error? Is there any way to derive $w=12$ from the starting problem? Thank you for your time.
I don't know where you got "$h \cdot 4h \cdot 12h = 768$" from.
The correct deductions from what you're given are $l = 4h$ and $h = \frac 13 w$, and hence $l = \frac 43 w$. Now $lwh = \frac 49 w^3 = 768$, and it follows that $w = \sqrt[3]{1728} = 12$.