Answer
(a) the ratio $\frac{h}{r} = \sqrt 2 $ gives $V = \frac{{\sqrt 2 }}{3}\pi {r^3}$ a maximum value subject to the given surface area.
(b) the ratio $\frac{h}{r} = \sqrt 2 $ gives $S = \sqrt 3 \pi {r^2}$ a minimum value subject to the given volume.
(c) a cone with given volume and maximum surface area does not exist.
Work Step by Step
For a right-circular cone of radius $r$ and height $h$ we have
- the surface area: $S = \pi r\sqrt {{r^2} + {h^2}} $
- the volume: $V = \frac{1}{3}\pi {r^2}h$
(a) In this case we have the volume $V\left( {r,h} \right) = \frac{1}{3}\pi {r^2}h$ subject to the constraint ${S_0} = \pi r\sqrt {{r^2} + {h^2}} $, where ${S_0}$ is a given constant.
Write the constraint equation: $g\left( {r,h} \right) = \pi r\sqrt {{r^2} + {h^2}} - {S_0} = 0$.
Using Theorem 1, the Lagrange condition $\nabla V = \lambda \nabla g$ yields
$\left( {\frac{2}{3}\pi rh,\frac{1}{3}\pi {r^2}} \right) = \lambda \left( {\pi \sqrt {{r^2} + {h^2}} + \frac{{\pi {r^2}}}{{\sqrt {{r^2} + {h^2}} }},\frac{{\pi rh}}{{\sqrt {{r^2} + {h^2}} }}} \right)$
$\left( {\frac{2}{3}\pi rh,\frac{1}{3}\pi {r^2}} \right) = \lambda \left( {\frac{{\pi \left( {2{r^2} + {h^2}} \right)}}{{\sqrt {{r^2} + {h^2}} }},\frac{{\pi rh}}{{\sqrt {{r^2} + {h^2}} }}} \right)$
So, the Lagrange equations are
$\frac{2}{3}rh = \lambda \frac{{2{r^2} + {h^2}}}{{\sqrt {{r^2} + {h^2}} }}$
$\frac{1}{3}{r^2} = \lambda \frac{{rh}}{{\sqrt {{r^2} + {h^2}} }}$
We may assume that $r \ne 0$ and $h \ne 0$, so
$\lambda = \frac{{2rh\sqrt {{r^2} + {h^2}} }}{{3\left( {2{r^2} + {h^2}} \right)}} = \frac{{{r^2}\sqrt {{r^2} + {h^2}} }}{{3rh}}$
This equation can be simplified to $\frac{{2rh}}{{2{r^2} + {h^2}}} = \frac{r}{h}$:
$2r{h^2} = r\left( {2{r^2} + {h^2}} \right)$
$2{h^2} = 2{r^2} + {h^2}$
${h^2} = 2{r^2}$
$h = \sqrt 2 r$
Thus, the critical point occurs when $\frac{h}{r} = \sqrt 2 $.
Using $h = \sqrt 2 r$, the extreme value of $V$ subject to the constraint ${S_0} = \pi r\sqrt {{r^2} + {h^2}} $ is
$V = \frac{1}{3}\pi {r^2}\left( {\sqrt 2 r} \right) = \frac{{\sqrt 2 }}{3}\pi {r^3}$
Since $h>0$, for a given ${S_0}$ we cannot increase $r$ without violating the constraint. Hence, we conclude that the ratio $\frac{h}{r} = \sqrt 2 $ gives $V = \frac{{\sqrt 2 }}{3}\pi {r^3}$ a maximum value subject to the constraint ${S_0}$.
(b) In this case we have the surface area $S\left( {r,h} \right) = \pi r\sqrt {{r^2} + {h^2}} $ subject to the constraint ${V_0} = \frac{1}{3}\pi {r^2}h$, where ${V_0}$ is a given constant.
Write the constraint equation: $g\left( {r,h} \right) = \frac{1}{3}\pi {r^2}h - {V_0} = 0$.
Using Theorem 1, the Lagrange condition $\nabla S = \lambda \nabla g$ yields
$\left( {\pi \sqrt {{r^2} + {h^2}} + \frac{{\pi {r^2}}}{{\sqrt {{r^2} + {h^2}} }},\frac{{\pi rh}}{{\sqrt {{r^2} + {h^2}} }}} \right) = \lambda \left( {\frac{2}{3}\pi rh,\frac{1}{3}\pi {r^2}} \right)$
$\left( {\frac{{2{r^2} + {h^2}}}{{\sqrt {{r^2} + {h^2}} }},\frac{{rh}}{{\sqrt {{r^2} + {h^2}} }}} \right) = \lambda \left( {\frac{2}{3}rh,\frac{1}{3}{r^2}} \right)$
So, the Lagrange equations are
$\frac{{2{r^2} + {h^2}}}{{\sqrt {{r^2} + {h^2}} }} = \lambda \left( {\frac{2}{3}rh} \right)$
$\frac{{rh}}{{\sqrt {{r^2} + {h^2}} }} = \lambda \left( {\frac{1}{3}{r^2}} \right)$
We may assume that $r \ne 0$ and $h \ne 0$, so
$\lambda = 3\frac{{2{r^2} + {h^2}}}{{2rh\sqrt {{r^2} + {h^2}} }} = 3\frac{h}{{r\sqrt {{r^2} + {h^2}} }}$
This equation can be simplified to $\frac{{2{r^2} + {h^2}}}{{2h}} = h$.
$2{r^2} + {h^2} = 2{h^2}$
${h^2} = 2{r^2}$
$h = \sqrt 2 r$
Thus, the critical point occurs when $\frac{h}{r} = \sqrt 2 $.
Using $h = \sqrt 2 r$ we obtain the extreme value of $S$ subject to the constraint ${V_0} = \frac{1}{3}\pi {r^2}h$:
$S = \pi r\sqrt {{r^2} + 2{r^2}} = \sqrt 3 \pi {r^2}$
For a given ${V_0}$ we may increase $r$ and at the same time decrease $h$ such that ${V_0}$ is constant. Hence, we conclude that the ratio $\frac{h}{r} = \sqrt 2 $ gives $S = \sqrt 3 \pi {r^2}$ a minimum value subject to the constraint ${V_0}$.
(c) In part (b) we obtain the minimum value of $S$ on the constraint ${V_0} = \frac{1}{3}\pi {r^2}h$:
$S = \sqrt 3 \pi {r^2}$
As we increase $r$ and at the same time decrease $h$, we can keep ${V_0}$ constant. However, as $r$ increases, $S$ increases. Hence, we conclude that there is no maximum value of $S$ on the constraint. Hence, a cone with given volume ${V_0}$ and maximum surface area does not exist.
