0
$\begingroup$

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 $f '(x) = 0$.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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:

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

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy