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"32. A hemisphere of radius $r$ rests with its plane face in contact with a horizontal plane. Find the least possible curved surface area of a right circular cone with its base on the plane and enclosing the hemisphere.

Prove that the same cone will also have the least possible volume, and that this volume is $\frac{1}{2}\pi{}r^{3}\sqrt{3}$."

(Background info: This question occurs at the end of a chapter on basic differentiation.)

What I know is that the curved surface area of the cone is $\pi{}rl$ (where $r$ is the radius of the base of the cone, not necessarily the radius of the hemisphere in the question.)

That's about it, so I decided to draw them in cross-section.

enter image description here

I think that the curved surface area will be lowest if the two sides of the cone are flat against the hemisphere. In other words, the sides of the cone will be tangents to the circle which is the cross-section of the hemisphere. If you see what I mean? So that's how I drew it in the diagram. That means that the angle between the radius of the circle and the side of the cone will be 90 degrees, as shown. So I've got a right-angled triangle there. I know that the two red angles will be the same, because it's a right cone. I know also that $l^{2} = (r + y)^{2} + (r + x)^{2}$. The problem is, all I've managed to do is introduce more unknowns which I do not believe I have enough information to solve. I suspect there's a trick, but it's been so long since I did this kind of maths that I'm completely at sea.

Any hints?

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  • $\begingroup$ I calculus allowed? $\endgroup$ Commented Mar 13, 2015 at 18:06
  • $\begingroup$ Calculus is definitely allowed, I just had no idea how to apply it to this kind of question, so approached it as a trig question. $\endgroup$
    – Au101
    Commented Mar 13, 2015 at 18:07

3 Answers 3

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Label the vertex point, tangent point, cone max radius point and center taken anticlockwise as $ V,T,B , O$ respectively .

It is convenient to take height $ h = VO $.

Verify the distances from trigonometry of right angled triangles.

$ VT = \sqrt {(h^2-r^2)}$

For the right angled triangle $ OVTB$ .. triangles $ OTV, VTO $ are similar; so $ TO^2 = TV \cdot TB $

$ TB = \dfrac{r^2}{\sqrt {(h^2-r^2)} }$

$ BO =\dfrac{h\cdot r}{\sqrt {(h^2-r^2)}} $

$ BT =\dfrac{ r^2}{\sqrt {(h^2-r^2)}} $

Slant area = $ \pi\cdot BO \cdot VB = \pi\cdot r\cdot h^3/ (h^2-r^2) $

Differentiate and simplify to get

$ h/r = \csc \alpha = \csc TVO = \sqrt 3 $

$ \alpha = ~ 35.25 \; degrees $

Volume of cone $ V = \pi\cdot BO^2 \cdot h/3 $

leads to the desired result.

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  • $\begingroup$ Sorry, could you clarify why $TB = \dfrac{r^{2}}{\sqrt{(h^{2} - r^{2})}}$ $\endgroup$
    – Au101
    Commented Mar 13, 2015 at 23:36
  • $\begingroup$ Added explanation of similar triangles OTV, VTO as above. $\endgroup$
    – Narasimham
    Commented Mar 14, 2015 at 5:44
  • $\begingroup$ Okay thanks, I'm with you now as far as your equation for slant area. When I differentiate that, though, I just get: $\dfrac{d}{dr}\left(\dfrac{\pi{}h^{3}r}{h^{2} - r^{2}}\right) = \dfrac{\pi{}h^{3}(h^{2} + r^{2})}{(h^{2} - r^{2})^{2}}$ $\endgroup$
    – Au101
    Commented Mar 14, 2015 at 19:35
  • $\begingroup$ @Au101: You should get $ 3 r^2 = h^2 $, by treating $h$ as a variable and $r$ as constant. $\endgroup$
    – Narasimham
    Commented Mar 14, 2015 at 19:42
  • $\begingroup$ But do I not have $f(r) = \frac{g(r)}{h(r)}$? So I must use the quotient rule: $\dfrac{d}{dr}\left(\dfrac{\pi{}h^{3}r}{h^{2} - r^{2}}\right) = \dfrac{(h^{2} - r^{2})\pi{}h^{3} - \pi{}h^{3}r(-2r)}{(h^{2} - r^{2})^{2}}$. I don't think that will simplify to $3r^{2} = h^{2}$ will it? Have I missed something? $\endgroup$
    – Au101
    Commented Mar 14, 2015 at 19:48
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Hint

Let the sphere's radius be $r$, and the cone's radius $r+x$. By triangle similarity, you have that:

$$\frac{r+x}{r}=\frac{l}{r+y}$$ Plugging that in to your equation, we get that: $$l^2 = (r+x)^2 + (r+y)^2 = \frac{(r+x)^2(r+y)^2}{r^2}$$ You can then solve for $y$ in terms of $x$, leading to the area being only a function of $x$.

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I am a 10th standard student and i am not well with these equations.

I had a very bad experience By attending this question in my thslc examination

Anyway, let'S make this simple for My low level friends("Just like my level").

--/ Let'S make this simple using an example. --/

Name the triangle ABC. 'A' at left, 'B' at Right,and 'C' at top.

    Perpendicular fro C to AB as a point 'D'

Also the point - Radius to the tangent as 'P'

                                Height         =20
                                Slant height=25

Considering Triangle ADC We can simply find radius of cone using pythogorean theorem Therefore i found out radius=15 We need to find the radius of the hemisphere. From the figure we know that 'r' oF sphere = PD

WE JUST KNOW ALL SIDES OF triangle ADC "Consider triangle ADC and ADP" They are similar.Just think it.You will get how

     Using similarity therom And  relation

AC/AD  =  CD/PD

Ie. 25/15 = 20/PB

BY JUST CROSS MULTIPLYING WE GET PB= 12

We found out the radius of the hemisphere.Rest is in your hand

Thank you :)

Sorry i need at least 10 Reputation to post image of this. Also sorry for any mistakes.

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