Bottles are either packed in boxes of 6 OR 12.The number of small boxes at least the half number of big boxes.If 240 bottles need to be packed,…

Question

Bottles are either packed in boxes of 6 OR 12. The number of small boxes must atleast be half the number of big boxes. If 240 bottles need to be packed, what's the minimum mumber of boxes needed?

This question seemed really simple to me. I thought using many big boxes would cause a little amount of boxes to be used so $\frac{240}{12} = 20$ boxes to be used the solutions said that the answer is 24? Why is this? I thought, in Mathematics the word or means 1 condition has to be true OR BOTH CONDITIONS CAN BE TRUE. Where did I go wrong?

Answer 1

You have not taken into account that the number of small boxes must be at least half the number of big boxes. You have no small boxes so this condition has not been satisfied in your answer.

You are correct in thinking that you should have as few small boxes as possible. If you let $p =$ the no. of small boxes then $6p$ bottles will be stored in these small boxes. There will be $240-6p$ bottles left to store in the big boxes and you will use $\frac{(240-6p)}{12}$ in the big boxes.

Since the number of small boxes must be at least half the number of big boxes, $p$ is greater than or equal to $\frac{0.5(240-6p)}{12}$ Solving this inequality results in $p$ being greater than or equal to $8.$ But you must have as few small boxes as possible so you must have $8$ small boxes - recall that $p$ is the number of small boxes. You can then use this to work out that you need $16$ big boxes by substituting $p$ into $\frac{(240-6p)}{12}.$

Hence you need a total of $24$ boxes altogether.

Answer 2

Suppose that you are using $x$ small boxes.

Then you are also using $(240-6x)/12$ large boxes.

Let $f(x)$ denote the total number of boxes used, which is $x+(240-6x)/12=x/2+20$.

$f(x)$ is monotonously increasing, so you should basically choose the minimal possible value of $x$.

But due to the constraint $x \geq ((240-6x)/12)/2 \rightarrow x \geq 8$, the minimal possible value of $x$ is $8$.

So use 8 small boxes and 16 large boxes.