# Minimize surface area with fixed volume [square based pyramid]

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 $$f '(x) = 0$$.

• So, given a fixed volume value, you want to maximize a square-based pyrmaid surface area? – NoChance Jun 6 '19 at 0:21
• No, minimize. I want the surface area to be as small as possible for the fixed volume. – user679930 Jun 6 '19 at 0:23
• 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 $(a)$. – NoChance Jun 6 '19 at 0:39

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:

$$\text{H}=\frac{3\cdot21^\frac{2}{3}}{2^\frac{2}{3}\cdot5^\frac{1}{3}}\approx8.41248\space\text{cm}\tag8$$