The question is:
A cone is to have a slant height of 22 cm. Find the vertical height h if its volume is to be a maximum.
Volume(cone) = ⅓πr²h
Can anyone help me? I have a little bit of idea as to how to approach this but my execution fails.
Thanks.
[Math] Calculate vertical height of cone if its volume is to be a maximum.
calculus
Related Solutions
A vertical cross section of the solid gives two similar triangles, one inside the other. The base of the larger triangle is $16$ and its height is $12$; the base of the smaller triangle is $12$ and its height is, say, $h$.
$h$/$6$ = $12$/$8$
which gives $h$ = $9$.
Now the volume can be found using the formula stated in the problem.
Assume that $r$= radius of cylinder = radius of cone = height of cylinder = height of cone.
Hence we can calulate the following:
- Volume of cylinder = volume of water = $\pi r^3$
- Volume of cone = $\frac13 \pi r^3$
- Volume of container = total volumes of cylinder and conee = $\frac 43 \pi r^3$
- Volume of air in container = $\frac 13 \pi r^3$
We know that
- (from above) the volume of water is more than half the volume of container
- container is symmetrical about its longitudinal axis
hence when container is lying on its side, the level of water must be greater than the midheight, $r$, of the container (on its side).
Let the level of water in the container on its side be $r+a$ where $0<a<1$.
(a) Cylinder
Volume of air is the area of a circular segment multiplied by its height $r$, i.e. $$\begin{align} V_1&=\frac 12 r^3 \ (2\alpha - \sin 2\alpha)\cdot r\qquad \text{where $\cos\alpha=\frac ar$}\\ &=\frac 12 r^3\ (2\alpha-2\sin\alpha\cos\alpha)\\ &=r^3\left[\cos^{-1}\left(\frac ar\right) -\left(\frac ar\right)\sqrt{1-\left(\frac ar\right)^2}\ \ \right]\\ &=r^3(\cos^{-1}\lambda-\lambda\sqrt{1-\lambda^2})\qquad \text{where $\lambda=\frac ar$}\\ &=\frac{r^3}3\left[3\cos^{-1}\lambda-3\lambda\sqrt{1-\lambda^2}\right] \end{align}$$
(b) Cone
Calculate the the volume of air in the cone by integrating circle segments of infinitesimally small thickness and of different radius $u$, as $u$ varies from $r$ (the part adjacent to the cylinder) to $a$ (edge of container).
Area of circle segment with radius $u$ is given by $$\frac 12 u^2 \ (2\phi - \sin 2\phi) = u^2\ \left[ \cos^{-1} \left(\frac au\right) - \frac au \sqrt{1-\left(\frac au \right)^2} \;\right]\qquad \text{where $\cos\phi=\frac au$}\\$$
Volume of air in cone is given by integrating this from $u=r$ to $u=a$, i.e. $$\begin{align} V_2&=\int_r^a u^2\ \left[ \cos^{-1} \left(\frac au\right) - \frac au \sqrt{1-\left(\frac au \right)^2}\;\right] du\\ &=\int_r^a u^2\cos^{-1} \left(\frac au\right) -a\sqrt{u^2-a^2} \; du\\ &=a^2 \int_r^a \left(\frac ua \right)^2\cos^{-1} \left(\frac au\right) -\sqrt{\left(\frac ua\right)^2-1} \; \; du\ \end{align}$$
Put $$\begin{align} \frac au = \cos\phi \Rightarrow u&=a\sec\phi\quad \Rightarrow \sec\phi=\frac ua\\ du&=a\sec\phi\tan\phi \; d\phi\\ u=r \Rightarrow \phi&=\cos^{-1}\frac ar=\cos^{-1}\lambda\\ u=a \Rightarrow \phi&=\cos^{-1}1=0 \end{align}$$
This gives $$\begin{align} V_2&=a^2\int_{\cos^{-1}\lambda}^{0}\left[(\sec^2\phi) \cdot \phi-\sqrt{\sec^2\phi-1}\right]\ a\sec\phi\tan\phi \; d\phi\\ &=a^3\int_{\cos^{-1}\lambda}^0 \phi\sec^3\phi\tan\phi-\sec\phi\tan^2\phi \; d\phi\\ &=a^3\int_{\cos^{-1}\lambda}^0 \phi\sec^3\phi\tan\phi-\sec^3\phi +\sec\phi \; d\phi\\ &=\cdots \text{(and after some tedious integration)}\cdots\\ &=\frac {a^3}3 \left[\phi\sec^3\phi-2\sec\phi\sqrt{\sec^2\phi-1}+\ln(\sec\phi+\ sqrt{\sec^2\phi-1})\right]_0^{\cos^{-1}\lambda}\\ &=\frac {a^3}3\left[ \left(\cos^{-1}\frac au\right)\left(\frac ua\right)^3-2\frac ua\sqrt{\left(\frac ua\right)^2-1}+\ln\left(\frac ua+\sqrt{\left(\frac ua\right)^2-1}\right)\right]_{u=a}^{u=r}\\ &=\frac {a^3}3 \left[ \left(\cos^{-1} \frac ar\right)\left(\frac ra\right)^3-2\frac ra \sqrt{\left(\frac ra\right)^2-1}+\ln\left(\frac ra+\sqrt{\left(\frac ra\right)^2-1}\right)\right]\\ &=\frac {a^3}3 \left[ \frac 1{\lambda^3} \cos^{-1}\lambda -\frac 2\lambda \sqrt{\frac 1{\lambda^2}-1}+\ln\left(\frac 1\lambda+\sqrt{\frac 1{\lambda^2}-1}\right) \right]\\ &=\frac {r^3}3 \left[ \cos^{-1} \lambda-2\lambda\sqrt{1-\lambda^2}+\lambda^3\ln\left(\frac 1\lambda +\sqrt{\frac 1{\lambda^2}-1}\right)\right] \end{align}$$
(c) Total
Total volume of air in container is $$\begin{align} V_1+V_2&=\frac {\pi r^3}3\\ \frac{r^3}3\left[4\cos^{-1}\lambda-5\lambda\sqrt{1-\lambda^2}+\lambda^3\ln\left(\frac 1\lambda (1 + \sqrt{1-\lambda^2})\right)\right]&=\frac {\pi r^3}3\\ 4\cos^{-1}\lambda-5\lambda\sqrt{1-\lambda^2}+\lambda^3\ln\left(\frac 1\lambda (1 + \sqrt{1-\lambda^2})\right)-\pi&=0\\ \lambda &\approx 0.36876\\ a=\lambda r &\approx 0.36876 r \end{align}$$
Hence, height of water in container on its side is $$r+a=(1+\lambda)r\approx 1.36876r \qquad \blacksquare$$
NB:
1. See this for a useful guide to the messy integration and the volume of a vertical cut of a cone.
2. See this for a useful guide for integrating $\sec^3\theta$.
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Best Answer
By pythagorus theorem,
$r^2+h^2=(22)^2$
$ V=π[(22)^2-h^2]h$
(for extreme points)
$\frac{dV}{dh}= π(22)^2-3h^2π=0$
→$$h=\frac{22}{√3}$$
$$\frac{d^2 V}{dx^2}=-6πh$$
$$\frac{d^2 V}{dx^2}<0$$
Ans
Volume is maximum at $$h=\frac{22}{√3}$$