[Math] Optimization of cake pan volume from area of pan

derivativesoptimization

It was difficult to accurately word this question, so hopefully a bit of context will clear that up.

Context:

I have a cake dish that is made by cutting out squares from the corners of a 25cm by 40 cm rectangle of tin. 

                          40cm

                 _____________________
                |_|                 |_|
       25cm     |                     |
                |                     |
                |_                   _|
                | |_________________|_|

A 3D cake tin is made by folding the edges once the squares have been cut away.

What size squares must be cut out to produce a cake dish of maximum volume?

My working:

  1. I know that the area of the pan without the squares will be:

    (40 – 2X) * (25-2Y) = Volume

  2. But that's about all I can wrap my head around. I know that the pieces cut off are sqaures, so they will have the same width and length. But that's all I can think of doing…

How exactly can I find what size square will produce a maximum volume for the container?

I'm pretty terrible at math, and I know it looks like I've done nothing to try and solve this. But I honestly am at a bit of a loss.

Best Answer

Hint: Suppose we remove $x\times x$ squares. The "finished" pan will have length $40-2x$, width $25-2x$, and depth $x$. So its volume $V(x)$ is given by $$V(x)=(40-2x)(25-2x)(x).$$ You want to choose $x$ that maximizes $V(x)$. Note that we will need $0\lt x$ and $2x\lt 25$.

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