Question

What proportions will maximize the volume of a projectile in the form of a circular cylinder with one conical end and one flat end, if the surface area is given?

Short answer

The resultant answer is to$r:l:s=\sqrt{5}:(1+\sqrt{5}):3.$

Step-by-step solution

Given data

A circular cylinder with conical end.

Concept of Partial differential equation

A partial differential equation consists of two or more independent variables, an unknown function (depending on those variables), and partial derivatives of the unknown function with respect to the independent variables.

Simplify the expression

From the given information, it is a typical optimization problem with constraint, where the area $\left(A\right)$is kept constant and need to find the proportions that will maximize this volume $\left(V\right)$.

So, will use Lagrange multiplier method is given below:

$$\begin{gathered} A=A_{\text{cylinder}}+A_{\text{cone}}=\text{const} \\ A=2\pi rl+\pi r^2+\pi rs \\ V=V_{\text{cylinder}}+V_{\text{cone}} \\ V=\pi r^{2}l+\frac{1}{3}\pi r^{2}h \end{gathered}$$ ……. (1)

Where, $h$is the height of the cone.

$V=\pi r^{2}l+\frac{1}{3}\pi r^2{\left(s^2-r^2\right)}^{1/2}$

Now, invoke the Lagrange multiplier method.

$F=\pi r^{2}2l+\frac{1}{3}\pi r^2{\left(s^2-r^2\right)}^{1/2}+\lambda \left(2\pi rl+\pi r^2+\pi rs\right)$

Take the partial derivative with respect to $r,s$, and $l$.

$$\begin{gathered} \frac{\partial F}{\partial r}=2\pi rl+\frac{2\pi r}{3}\left({}_{-}{}^{s^2}-\frac{\pi r^2}{3}\left(\frac{1}{2}\times 2r\times {\left(s^2-r^2\right)}^{1/2}\right)+\lambda \pi (2r+2l+s)=0\right. \\ \frac{\partial F}{\partial r}=2rl+\frac{2r}{3}\left(s_{-}^2-\frac{r^3}{3}\times \left({}_{-}{}^2+\lambda (2r+2l+s)=0\right.\right. \end{gathered}$$ ……. (2)

Calculate further as follows:

$\begin{array}{rcl}\frac{\partial F}{\partial s}& =& \frac{1}{3}\pi r^2\left(\frac{1}{2}\times {\left(s^2-r^2\right)}^{-1/2}\times 2s\right)+\lambda \pi r\\ & =& 0\\ \frac{\partial F}{\partial l}& =& \frac{r^{2}s}{3}\left(s_{-}^2+\lambda r=0\right.\end{array}$ …….(3)

Similarly, calculate further.

$\begin{array}{rcl}\frac{\partial F}{\partial l}& =& \pi r^2+\lambda \left(2\pi r\right)\\ \frac{\partial F}{\partial l}& =& 0\\ \frac{\partial F}{\partial l}& =& r(r+2\lambda )\\ \frac{\partial F}{\partial l}& =& 0\end{array}$

Derive the expression

Now by three partial derivative equations and the original constraints equation, will be solve for unknown proportions.

By using equation (4) $\lambda =-\frac{r}{2}$

Substitute for $\lambda$in equation (3).

Now substitute in equation (2)

So, $\sqrt{5}l=\left(1+\sqrt{5}\right)r$.

Thus, the ratio between the three quantities $r:l:s=\sqrt{5}:\left(1+\sqrt{5}\right):3$.