# Question on surface area of cone

• Aug 16th 2007, 07:18 PM
acc100jt
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 $\displaystyle cm^3$ of chocolate.

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

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

c) The height of the cone in (b) can be expressed as $\displaystyle h=\sqrt{ar}$, 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??
• Aug 17th 2007, 12:17 AM
ticbol
Quote:

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 $\displaystyle cm^3$ of chocolate.

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

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

c) The height of the cone in (b) can be expressed as $\displaystyle h=\sqrt{ar}$, 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
• Aug 17th 2007, 02:23 AM
acc100jt
Thanks a lot.
The answer given in the exercise book for part (c) is 2, i.e. $\displaystyle a=2$.
But I couldn't get $\displaystyle a=2$, now you have confirmed my answer is correct. Thanks!!!
• Aug 17th 2007, 03:40 AM
Soroban
Hello, acc100jt!

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

Quote:

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