**Volume optimization of a cuboid**

This is an extension of the Nrich task which is currently live – where students have to find the maximum volume of a cuboid formed by cutting squares of size x from each corner of a 20 x 20 piece of paper. I’m going to use an n x 10 rectangle and see what the optimum x value is when n tends to infinity.

First we can find the volume of the cuboid:

Next we want to find when the volume is a maximum, so differentiate and set this equal to 0.

Next we use the quadratic formula to find the roots of the quadratic, and then see what happens as n tends to infinity (i.e we want to see what the optimum x values are for our cuboid when n approaches infinity). We only take the negative solution of the + – quadratic solutions because this will be the only one that fits the initial problem.

Next we try and simplify the square root by taking out a factor of 16, and then we complete the square for the term inside the square root (this will be useful next!)

Next we make a u substitution. Note that this means that as n approaches infinity, u approaches 0.

Substituting this into the expression gives us:

We then manipulate the surd further to get it in the following form:

Now, the reason for all that manipulation becomes apparent – we can use the binomial expansion for the square root of 1 + u^{2} to get the following:

Therefore we have shown that as the value of n approaches infinity, the value of x that gives the optimum volume approaches 2.5cm.

So, even though we start with a pretty simple optimization task, it quickly develops into some quite complicated mathematics. We could obviously have plotted the term in n to see what its behavior was as n approaches infinity, but it’s nicer to prove it. So, let’s check our result graphically.

As we can see from the graph, with n plotted on the x axis and x plotted on the y axis we approach x = 2.5 as n approaches infinity – as required.

**An m by n rectangle.**

So, we can then extend this by considering an n by m rectangle, where m is fixed and then n tends to infinity. As before the question is what is the value of x which gives the maximum volume as n tends to infinity?

We do the same method. First we write the equation for the volume and put it into the quadratic formula.

Next we complete the square, and make the u substitution:

Next we simplify the surd, and then use the expansion for the square root of 1 + u^{2}

This then gives the following answer:

So, we can see that for an n by m rectangle, as m is fixed and n tends to infinity, the value of x which gives the optimum volume tends to m/4. For example when we had a 10 by n rectangle (i.e m = 10) we had x = 2.5. When we have a 20 by n rectangle we would have x = 5 etc.

And we’ve finished! See what other things you can explore with this problem.

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