We're doing a calculus contest/project in school. In short, we need to see who can come up with the most creative modification to an existing container. The fixed volume I have to work with is $99.225\ \mathrm{cm}^3$.

I'm trying to use substitution to solve in my surface area, in which my height ($h$) is equal to $\dfrac{297.675}{x^2}$. My side length is represented by $x$.

Inserting this into the surface area of a square based pyramid, I get:

$$f(x)= 2x \sqrt{\frac{88610.40563}{x^4}+\frac{x^2}{4}} +x^2$$

And this is where I get stuck. I don't know how to proceed with this equation to continue simplifying and ultimately determining the first derivative of this equation, so that I can find the minimum value of $x$ when $f '(x) = 0$.

  • $\begingroup$ So, given a fixed volume value, you want to maximize a square-based pyrmaid surface area? $\endgroup$ – NoChance Jun 6 '19 at 0:21
  • $\begingroup$ No, minimize. I want the surface area to be as small as possible for the fixed volume. $\endgroup$ – user679930 Jun 6 '19 at 0:23
  • $\begingroup$ If you use the volume formula $v=a^2\frac{h}{2}$ and solve for $a$, you can determine the surface area. You can't min/max the surface area given height and base length $(a)$. $\endgroup$ – NoChance Jun 6 '19 at 0:39

Well, the volume of a square pyramid is given by:


Where the base length is given by $\text{L}$ and perpendicular height is given by $\text{H}$.

A right square pyramid with base length $\text{L}$ and perpendicular height $\text{H}$ has surface area of:


With a given volume we can solve equation $(1)$ for $\text{H}$:


Substitute equation $(3)$ into equation $(2)$ gives:


Now, we need to solve:


And it gives:


So, when $\mathcal{V}=\frac{99225}{1000}\space\text{cm}^3$, it gives for $\text{L}$:


And so the height is given by:


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