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I am trying to come up an equation for the maximum volume for a prism with a base of a regular polygon, given that the surface area is 100 units squared. I ended up with these two equations, but I'm not sure how to optimize as there are three variables for only two equations. Link to equations: https://docs.google.com/document/d/1-qH5ErluVdQNmIDeXBGF0z-jMB6BYLI4yNb4EkmitUc/edit?usp=sharing

$n$- number of sides

$s$- length of side

$h$- height of prism.

$$ SA = \frac{ns^2}{2 \tan(\frac{\pi}{n})} + nhs = 100. $$ $$ Vol = \frac{200s \tan \frac{\pi}{n} - ns^3}{8 \tan^2 \frac{\pi}{n}} $$

It is similar to the investigation in this website(https://ibmathsresources.com/2017/05/21/optimization-of-area-an-investigation/), but with prisms instead of planes.

John Hughes
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mbzht
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  • It's generally recommended to put the equations directly into your question. For help, see https://math.meta.stackexchange.com/questions/1773/do-we-have-an-equation-editing-howto – David K Jan 13 '20 at 01:48
  • It's also helpful to say what each of your named variables is (number of sides, height of the prism, etc.) and to explain how you derived your equations. In your case the volume equation seems obscure. – David K Jan 13 '20 at 01:49
  • I've typed in your equations for you. I agree with David K that your volume equation seems peculiar. The $ns^3$ in the numerator should surely be $ns^2$ (assuming you're solving the first equation to find $h$ in terms of $n$ and $s$). To solve your system (once you get it right), you'll fix $n$, and use the surface-area equation to derive $h$ in terms of $s$; then you can write $Vol$ as a function of just $s$, differentiate, and find the max vol (in terms of $n$). Then you look at this as $n$ varies, and discover that it grows with $n$, so the best "number of sides" is infinite (I'm guessing). – John Hughes Jan 13 '20 at 02:11
  • I've tested my equation using specific s and h value from past calculations, and it seems to be correct. Could you aid me in understanding how to get the Volume equation in terms of only s. – mbzht Jan 13 '20 at 02:41

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Let's look at the case $n = 4, s = 1, h = 1/2$.

Then you've got a pyramid erected on a square. The area of the square is $s^2 = 1$. The altitude of each triangular side is $\frac{\sqrt{2}}{2}$. So the area of the four triangular sides is $$ 4 \cdot (\frac12 \cdot 1 \cdot \frac{\sqrt{2}}{2}) = \sqrt{2}. $$ So the total area of the pyramid is $1 + \sqrt{2}$.

Your area formula says that the surface area of the pyramid is $$ \begin{align} SA &= \frac{ns^2}{2 \tan(\frac{\pi}{n})} + nhs\\ &= \frac{4 \cdot 1^2}{2 \tan(\frac{\pi}{4})} + 4\cdot \frac12 \cdot 1\\ &= \frac{4 }{2 \cdot 1} + 4\cdot \frac12 \cdot 1\\ &= 2 + 2 = 4.\\ \end{align} $$ So... something's wrong. Until you get that area formula right, setting it to $100$ isn't going to yield anything good.

John Hughes
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