Hello, acc100jt!

This is how I did part (a) . . .

A manufacturer intends to use the least amount of material to make

a hollow right circular cone of radius $\displaystyle r$ cm and vertical height $\displaystyle h$ cm .

The cone must be able to contain 100 cm³ of chocolate.

a) Show that the amount of material used or the curved surface area

of the cone is: . $\displaystyle S\:=\:\frac{\sqrt{\pi^2 r^6+90000}}{r}$

The side view of the cone looks like this: Code:

r
* - - + - - *
\ | /
\ | /
\ |h / R
\ | /
\|/
*

$\displaystyle r$ is the radius of the cone, $\displaystyle h$ is the height of the cone.

The slant height of the cone is: .$\displaystyle R \:=\:\sqrt{r^2+h^2}$ .**[1]**

Since the volume is 100 cm³: .$\displaystyle V \:=\:\frac{1}{3}r^2h \:=\:100\quad\Rightarrow\quad h \:=\:\frac{300}{\pi r^2}$

Substitute into [1]: .$\displaystyle R \:=\:\sqrt{r^2 + \left(\frac{300}{\pi r^2}\right)^2} \:=\:\sqrt{r^2+\frac{90,000}{\pi^2r^4}} :=\:\sqrt{\frac{\pi^2r^6 + 90,000}{\pi^2r^4}}$

. . Hence: .$\displaystyle R\:=\:\frac{\sqrt{\pi^2r^6 + 90,000}}{\pi r^2} $ .**[2]**

The cone is formed from a sector of the circle

. . with radius $\displaystyle R$ and central angle $\displaystyle \theta$. Code:

* * *
* *
* *
* *
* *
* * *
* * θ * *
* * R
* * * *
* *
* *
* * *
Rθ

The area of the sector is: .$\displaystyle S \:=\:\frac{1}{2}R^2\theta$ .**[3]**

The arclength of the sector $\displaystyle (R\theta)$,

. . is the circumference of the base of the cone $\displaystyle (2\pi r)$.

Hence: .$\displaystyle R\theta \:=\:2\pi r\quad\Rightarrow \quad\theta \:=\:\frac{2\pi r}{R}$

Substitute into [3]: .$\displaystyle S \:=\:\frac{1}{2}R^2\left(\frac{2\pi r}{R}\right)\quad\Rightarrow\quad S \:=\:\pi rR$ .**[4]**

Substitute [2] into [4]: .$\displaystyle S \:=\:\pi r\cdot\frac{\sqrt{\pi^2r^6 + 90,000}}{\pi r^2} $

Therefore: .$\displaystyle \boxed{S \;=\;\frac{\sqrt{\pi^2r^6 + 90,000}}{r}}$