# Math Help - Question on surface area of cone

1. ## Question on surface area of cone

A manufacturer intends to use the least amount of material to make a hollow right circular cone of radius r cm and vertical height h cm . The cone must be able to contain 100 $cm^3$ of chocolate.

a) Show that the amount of material used or the curved surface area of the cone is $S=\frac{\sqrt{\pi^2 r^6+90000}}{r}$

b) Find the radius of the cone so that the amoint of material used will be the minimum. Give your answer in terms of $\pi$

c) The height of the cone in (b) can be expressed as $h=\sqrt{ar}$, where a is a constant. Find the value of a

I am able to do (a) and (b), but can't do part (c). Can anyone help??

2. Originally Posted by acc100jt
A manufacturer intends to use the least amount of material to make a hollow right circular cone of radius r cm and vertical height h cm . The cone must be able to contain 100 $cm^3$ of chocolate.

a) Show that the amount of material used or the curved surface area of the cone is $S=\frac{\sqrt{\pi^2 r^6+90000}}{r}$

b) Find the radius of the cone so that the amoint of material used will be the minimum. Give your answer in terms of $\pi$

c) The height of the cone in (b) can be expressed as $h=\sqrt{ar}$, where a is a constant. Find the value of a

I am able to do (a) and (b), but can't do part (c). Can anyone help??
This is good for practice.

Given: right circular cone, V=100 cu.cm, radius r, height h.
V = pi(h/3)(r^2) = 100
h = 300/(pi*r^2) --------(1)

a) The surface area S.
S = (length of inclined height)(circumference at mid-height of the cone) <--- using Pappus' Centroid Theorem.
S = (sqrt[r^2 +h^2])(2pi[r/2])
S = (sqrt[r^2 +90000/(pi^2 *r^4)])(pi*r)
S = (sqrt[pi^2 *r^6 +90000] /(pi*r^2))(pi*r)
S = (sqrt[pi^2 *r^6 +90000])/(r) -----------------shown.

----------------------------------------------------------------------
b) Find r when S is minimum.
So, the r when dS/dr = 0.

For simplicity, let's simplify S, let's put the denominator r into the radical.
S = sqrt[(pi^2 *r^4) +(90000/(r^2))]
dS/dr = (1/2)[4(pi^2)(r^3) -2(90000)/(r^3)] /sqrt[(pi^2 *r^4) +(90000/(r^2))]
Set that to zero,
0 = 2(pi^2)(r^3) -90000/(r^3)
90000 = 2(pi^2)(r^6)
r^6 = 45000/(pi^2)
r = [45000/(pi^2)]^(1/6)
r = 4.072103 cm.

In terms of pi,
r = (4.072103)(pi/pi) = 1.29619 pi -----------answer.

------------------------------------------------------------
c) If h in (b) above is sqrt(a*r), find "a".

h = 300/(pi*r^2) --------(1)
h = 300/(pi *[(4.072103)^2)]
h = 5.758825 cm.
So,
5.758825 = sqrt[a(4.072103)]
Square both sides, and then,
a = 33.164065 / 4.072103

3. Thanks a lot.
The answer given in the exercise book for part (c) is 2, i.e. $a=2$.
But I couldn't get $a=2$, now you have confirmed my answer is correct. Thanks!!!

4. 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 $r$ cm and vertical height $h$ 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: . $S\:=\:\frac{\sqrt{\pi^2 r^6+90000}}{r}$

The side view of the cone looks like this:
Code:
               r
* - - + - - *
\    |    /
\   |   /
\  |h / R
\ | /
\|/
*
$r$ is the radius of the cone, $h$ is the height of the cone.
The slant height of the cone is: . $R \:=\:\sqrt{r^2+h^2}$ .[1]

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

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

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

The cone is formed from a sector of the circle
. . with radius $R$ and central angle $\theta$.
Code:
              * * *
*           *
*               *
*                 *

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

The arclength of the sector $(R\theta)$,
. . is the circumference of the base of the cone $(2\pi r)$.
Hence: . $R\theta \:=\:2\pi r\quad\Rightarrow \quad\theta \:=\:\frac{2\pi r}{R}$

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

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

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