Optimization of a rectangular container

Question

A rectangular sheet of tinplate is $2k$ cm by $k$ cm. Four squares, each with sides $x$ cm, are cut from its corners. The remainder is bent into the shape of an open rectangular container. Find the value of $x$ which will maximize the capacity of the container?

I tried making the sides: $2k - 2x$, $x$, and $k - 2x$. Then multiplying all 3 together. After I do that I find the derivative of the expression and set it equal to 0 to look for its max and mins. I end up using the quadratic formula and end up with $\frac{k}{2} \pm \frac{k\sqrt{3}}{6}$. But the answer is:

I was wondering if someone could help guide me in the right way?

Answer 1

What you have is correct. From $\frac{k}{2} \pm \frac{k\sqrt{3}}{6}$, you can exclude $\frac{k}{2} + \frac{k\sqrt{3}}{6}$ as an answer since it is greater than $\frac{k}{2}$. Hence, you get that the critical value is $$\frac{k}{2} - \frac{k\sqrt{3}}{6} = \frac{k}{2}\left(1 - \frac{1}{\sqrt{3}}\right)$$

Answer 2

Your answer is almost right. First, notice that $$ \frac{k}{2} \frac{1}{\sqrt{3}} = \frac{k \sqrt{3}}{6} $$ so it is just a matter of deciding whether the $\pm$ sign should be $+$, $-$, or both to get a maximum.

Well, if you use the $+$ sign, then the cut $x > \frac{k}{2}$ so the folds in you diagram along the short size add up to more than the length of the side, which makes no sense. So the $-$ sign must be used.

If you said the answer was $$ \frac{k}{2} - \frac{k\sqrt{3}}{6}$$ then your answer would be equivalent to the one given, and correct.