Name: Increasing Functions
Uploaded: 2019-07-19
Duration: 3 min 59 s
Description: The GDP of India is either increasing or decreasing, depending on a variety of factors. Let’s understand what an increasing function is, by using derivatives...

Question

Show that the right circular cone of the least curved surface and given volume has an altitude equal to

\sqrt{2}

times the radius of the base.

Accepted Answer

Let $r$ and $h$ be the radius and height of a cone respectively
and $V$ and $S$ be the volume and curved surface area of cone respectively.

Given volume of cone is constant
Volume of cone, $V = \frac{1}{3} π {(radius)}^{2} (height)$
$\frac{3 V}{π} = r^{2} h$
$h = (\frac{3 V}{π}) \frac{1}{r^{2}}$
$h = \frac{k}{r^{2}}$ $(∵ \frac{3 V}{π} = constant (k))$ .............. $(1)$

We need to show Curved surface Area is least
and altitude $= \sqrt{2}$ times of radius of base
i.e., $h = \sqrt{2} r$

Now, Curved Surface Area of Cone $= π r l$
substitute, $l = \sqrt{h^{2} + r^{2}}$
$S = π r \sqrt{h^{2} + r^{2}}$
$S = π r \sqrt{{(\frac{k}{r^{2}})}^{2} + r^{2}}$
$S = π r \sqrt{\frac{k^{2} + r^{6}}{r^{4}}}$
$S = π \frac{r}{r^{2}} \sqrt{k^{2} + r^{6}}$
$S = π [\frac{\sqrt{k^{2} + r^{6}}}{r}]$
Since $S$ has square root, it will be difficult to differentiate. So, we take
$Z = S^{2}$
$Z = π^{2} [\frac{k^{2} + r^{6}}{r^{2}}]$
$Z = π^{2} [\frac{k^{2}}{r^{2}} + r^{4}]$
$Z = π^{2} [k^{2} r^{- 2} + r^{4}]$
since $S$ is positive,
$S$ is minimum if $S^{2}$ is minimum
So, we minimize $Z = S^{2}$

Differentiate $Z$ w.r.t. $r$
$\frac{d Z}{d r} = \frac{d}{d r} (π^{2} [k^{2} r^{- 2} + r^{4}])$
$\frac{d Z}{d r} = π^{2} [- 2 k^{2} r^{- 3} + 4 r^{3}]$
Putting $\frac{d Z}{d r} = 0$
$π^{2} [- 2 k^{2} r^{- 3} + 4 r^{3}] = 0$
$- 2 k^{2} r^{- 3} + 4 r^{3} = 0$
$4 r^{3} = 2 k^{2} r^{- 3}$
$4 r^{3} = \frac{2 k^{2}}{r^{3}}$
$4 r^{3} \times r^{3} = 2 k^{2}$
$4 r^{6} = 2 k^{2}$
$r^{6} = \frac{k^{2}}{2}$

Finding $\frac{d^{2} Z}{d r^{2}}$
$\frac{d Z}{d r} = π^{2} [- 2 k^{2} r^{- 3} + 4 r^{3}]$
Differentiate w.r.t $r$
$\frac{d^{2} Z}{d r^{2}} = \frac{d}{d r} [π^{2} [- 2 k^{2} r^{- 3} + 4 r^{3}]]$
$\frac{d^{2} Z}{d r^{2}} = π^{2} [- 2 k^{2} (- 3 r^{- 4}) + 4 (3 r^{2})]$
$\frac{d^{2} Z}{d r^{2}} = π^{2} [6 k^{2} r^{- 4} + 12 r^{2}]$
Since R.H.S is a positive value,
$∴ \frac{d^{2} Z}{d r^{2}} > 0$ for $r^{6} = \frac{k^{2}}{2}$
Thus, Surface area is minimum when $r^{6} = \frac{k^{2}}{2}$ .

Now,
$h = \frac{k}{r^{2}}$ [from $(1)$ ]
$h r^{2} = k$
${(h r^{2})}^{2} = k^{2}$
$h^{2} r^{4} = k^{2}$
Putting value of $k^{2} = 2 r^{6}$
$h^{2} r^{4} = 2 r^{6}$
$h^{2} = \frac{2 r^{6}}{r^{4}}$
$h^{2} = 2 r^{2}$
$h = \sqrt{2 r^{2}}$
$h = r \sqrt{2}$
Hence, altitude equal to $\sqrt{2}$ times the radius of the base of right circular cone.
Hence it is proved.

Show that the right circular cone of the least curved surface and given volume has an altitude equal to $\sqrt{2}$ times the radius of the base.

Important Questions on Application of Derivatives

Learn and Prepare for any exam you want !

Show that the right circular cone of the least curved surface and given volume has an altitude equal to 2 times the radius of the base.

Important Questions on Application of Derivatives

Learn and Prepare for any exam you want !

Show that the right circular cone of the least curved surface and given volume has an altitude equal to $\sqrt{2}$ times the radius of the base.