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 $f '(x) = 0$.
Well, the volume of a square pyramid is given by:
$$\mathcal{V}=\frac{1}{3}\cdot\text{H}\cdot\text{L}^2\tag1$$
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:
$$\mathcal{A}=\text{L}^2+\text{L}\cdot\sqrt{\text{L}^2+\left(2\cdot\text{H}\right)^2}\tag2$$
With a given volume we can solve equation $(1)$ for $\text{H}$:
$$\mathcal{V}=\frac{1}{3}\cdot\text{H}\cdot\text{L}^2\space\Longleftrightarrow\space\text{H}=\frac{\mathcal{V}}{\frac{1}{3}\cdot\text{L}^2}=\frac{3\cdot\mathcal{V}}{\text{L}^2}\tag3$$
Substitute equation $(3)$ into equation $(2)$ gives:
$$\mathcal{A}=\text{L}^2+\text{L}\cdot\sqrt{\text{L}^2+\left(2\cdot\frac{3\cdot\mathcal{V}}{\text{L}^2}\right)^2}\tag4$$
Now, we need to solve:
$$\frac{\partial\mathcal{A}}{\partial\text{L}}=0\tag5$$
And it gives:
$$\text{L}^5\cdot\left(\text{L}+\sqrt{\text{L}^2+\frac{36\cdot\mathcal{V}^2}{\text{L}^4}}\right)-18\cdot\mathcal{V}^2=0\tag6$$
So, when $\mathcal{V}=\frac{99225}{1000}\space\text{cm}^3$, it gives for $\text{L}$:
$$\text{L}=\frac{3\cdot21^\frac{2}{3}}{2\cdot2^\frac{1}{6}\cdot5^\frac{1}{3}}\approx5.94852\space\text{cm}\tag7$$
And so the height is given by:
$$\text{H}=\frac{3\cdot21^\frac{2}{3}}{2^\frac{2}{3}\cdot5^\frac{1}{3}}\approx8.41248\space\text{cm}\tag8$$