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$r$ - radius of base
$h$ - height of cone
$l$ - slant height of cone
$V$ - Volume of cone
$A$ - surface area of cone

First of all I myself don't know what's the meaning of phrase "given surface area", since it is written given I would assume it to be constant. I need to maximize the volume: $$ V = \frac{\pi r^2 h}{3}, $$ $$ \frac{dV}{dr} = \frac {\pi}{3} \left(2rh + r^2 \frac{dh}{dr}\right). $$

To obtain critical points set $dV/dr = 0$: $$ 2rh + r^2 \frac{dh}{dr} = 0, $$ $$ \frac{dh}{dr} = -2\frac{h}{r}. $$ Now, $A = \pi r \left( r + \sqrt{h^2 + r^2} \right)$. Using $dA/dr = 0$ and $dh/dr = -2h/r$ we would get $$ h = \sqrt{8}r, $$ I have got the answer but how can I check whether it's a minimum volume or maximum volume.

For example if I put $h = \sqrt{8}r$ in the equation of volume we would get
$V = (\sqrt{8}/3) \pi r^3$, so if I double differentiate it $d^2V/dr^2 = 2\cdot \sqrt{8} \cdot \pi \cdot r^2 $, no matter what value or $r$ I put I always get a positive value which means $h = \sqrt{8}r$ represents a minimum volume.

Please spot out my mistake as I'm myself unable to do it.

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you seek an extremum of the function $$ f(r,h,l) = r^2h $$

subject to the imposed condition: $$ r^2 + rl = c $$ and the geometrical relation $$ h^2 + r^2 - l^2 = 0 $$

form the objective function: $$ L = r^2h + \lambda(r^2 + rl) + \mu(h^2 + r^2 - l^2) $$

for an extremum $$ \frac{\partial L}{\partial h} = r^2 + 2 \mu h = 0 \\ \frac{\partial L}{\partial r} = 2rh + \lambda(2r+l) + 2 \mu r = 0 \\ \frac{\partial L}{\partial l} = \lambda r - 2 l \mu = 0 $$ from which $$ 2 \mu = - \frac{r^2}h = \frac{\lambda r}l $$ giving $$ \lambda = 2\mu \frac{l}r $$ now, substituting in $(1)$ $$ 2h + 2\mu \left( 2l + \frac{l^2}r + r \right) = 0 $$ giving $$ 2h^2 = 2lr +r^2 + l^2 = 2lr + r^2 + r^2+h^2 $$ so, finally $$ h^2 = 2(lr + r^2) = \frac{2A}{\pi} $$

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  • $\begingroup$ Sir is it not l^2 -r^2-h^2 = 0 ? $\endgroup$ – Knight Mar 3 '19 at 9:40
  • $\begingroup$ using the expression that way round will merely cause a change of sign in $\mu$ which will not affect the outcome. the same (kind of) reason means it is enough to maximize $r^2 h$ rather than $\frac13\pi r^2h$. $\endgroup$ – David Holden Mar 3 '19 at 13:11
  • $\begingroup$ Sir what are lambda and mu ? What do they represent? $\endgroup$ – Knight Mar 3 '19 at 14:06
  • $\begingroup$ I have really liked your method. Only problem is you have missed some constants and it’s difficult for beginners like me to get convinced. $\endgroup$ – Knight Mar 3 '19 at 14:08
  • $\begingroup$ to find out more about the method, do a search on "lagrange multipliers two constraints". re the 'missing constants' you can work through the algebra above for a version with all the constants included, and you should see that this doesn't affect the outcome, but clutters up the working. it is true that we need experience to see which constants are irrelevant and may be ignored without loss of generality ;-) $\endgroup$ – David Holden Mar 3 '19 at 15:47
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Hint

For such a cone, we have

$$V=\frac 1 3 \pi r^2h \qquad \text{and} \qquad A=\pi r \left(r + \sqrt{r^2+h^2}\right)$$

Since $A$ is given, extract $h$ $$h=\frac{\sqrt{A} \sqrt{A-2 \pi r^2}}{\pi r}$$ Replace in $V$ and ... continue

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  • $\begingroup$ Thank you. Can you please tell me my mistake? $\endgroup$ – Knight Mar 3 '19 at 8:42
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    $\begingroup$ @adeshmishra. You have a relation between $h$ and $r$ since $A$ is fixed; this makes $h$ to be a function of $r$ ... what you did not exploit. $\endgroup$ – Claude Leibovici Mar 3 '19 at 9:21
  • $\begingroup$ Sir, I have sent you an email on the address you have given in your ‘about’. If you have time then please see my email. $\endgroup$ – Knight Jun 3 '19 at 9:41

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