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What would be the radius and the altitude of a right circular cone that circumscribes a sphere with a radius 8 cm if the volume of the cone is to be minimized?

Here is my rough sketch;

enter image description here

My idea is to write some characteristics of the cone as a function of the radius of the circle, minimize it with differential calculus and connect those characteristics to the base radius and height of the cone, however I'm stuck at step 1.

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Let $r,a$ be the radius and the altitude of a right circular cone respectively.

Then, since considering the vertical plane which cuts the cone in half gives us

$$BD=DE=r,AD=\sqrt{a^2+r^2},AE=\sqrt{a^2+r^2}-r$$ $$CB=CE=8,AC=\sqrt{8^2+\left(\sqrt{a^2+r^2}-r\right)^2},$$ we have $$AB=CB+AC\Rightarrow a=8+\sqrt{8^2+\left(\sqrt{a^2+r^2}-r\right)^2}.$$

Since this leads $$(a-8)^2=8^2+\left(\sqrt{a^2+r^2}-r\right)^2$$ $$\Rightarrow a^2-16a=(a^2+r^2)-2r\sqrt{a^2+r^2}+r^2$$ $$\Rightarrow r\sqrt{a^2+r^2}=r^2+8a\Rightarrow r^2(a^2+r^2)=(r^2+8a)^2$$ $$\Rightarrow r^2(a^2-16a)=64a^2\Rightarrow r^2=\frac{64a^2}{a^2-16a},$$ all you need to consider is the following function with only one variable for $a\gt 16$:

$$\frac{\pi r^2a}{3}=\frac{64\pi a^2}{3(a-16)}=f(a).$$ Since $$f'(a)=\frac{64\pi}{3}\cdot\frac{a(a-32)}{(a-16)^2},$$ the minimum is attained when $a=32,r=8\sqrt 2$.

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Let $y=AC$ and $r=BD$, so $h=y+8$.

By similar triangles, $\displaystyle\frac{r}{y+8}=\frac{8}{\sqrt{y^2-64}}$ $\;\;$so $\displaystyle r^2=\frac{64(y+8)^2}{y^2-64}=\frac{64(y+8)}{y-8}$ and

$V=\displaystyle\frac{1}{3}\pi r^2h=\frac{64\pi}{3}\cdot\frac{(y+8)^2}{y-8}$.

Then $\displaystyle V^{\prime}(y)=\frac{64\pi}{3}\cdot\frac{(y+8)(y-24)}{(y-8)^2}=0$ if $y=24$,

sor $r=\sqrt{\frac{64(32)}{16}}=\sqrt{64(2)}=8\sqrt{2}$ $\;\;$ and $\;\;h=24+8=32$.

(By the first-derivative test, $y=24$ gives a minimum.)


Alternate Solution:

Let $\theta=\angle ABD$, so $r=8\cot\frac{\theta}{2}=8(\csc\theta+\cot\theta)$ and $h=r\tan\theta=8(\csc\theta+\cot\theta)\tan\theta$.

Then $V=\frac{1}{3}\pi r^{2}h=\frac{512\pi}{3}(\csc\theta+\cot\theta)^3\tan\theta$,

so $V^{\prime}(\theta)=\frac{512\pi}{3}\left[(\csc\theta+\cot\theta)^{3}\sec^{2}\theta+3(\csc\theta+\cot\theta)^{2}(-\csc\theta\cot\theta-\csc^{2}\theta)\tan\theta\right]$

$\hspace{.6 in}=\frac{512\pi}{3}(\csc\theta+\cot\theta)^2\left[(\csc\theta+\cot\theta)\sec^{2}\theta-3(\csc\theta\cot\theta+\csc^{2}\theta)\tan\theta\right]=0$

$\implies(\csc\theta+\cot\theta)\sec^{2}\theta=3(\csc\theta\cot\theta+\csc^{2}\theta)\tan\theta$

$\implies\displaystyle\frac{1+\cos\theta}{\sin\theta\cos^{2}\theta}=\frac{3(1+\cos\theta)}{\cos\theta\sin\theta}\implies\cos\theta=\frac{1}{3}\implies\sin\theta=\frac{\sqrt{8}}{3}$,

so $\csc\theta+\cot\theta=\frac{1+\cos\theta}{\sin\theta}=\sqrt{2}\;\;$ and $\;\;r=8\sqrt{2},\;\;$ $h=r\tan\theta=(8\sqrt{2})\cdot\sqrt{8}=32$.

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