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    I'm having trouble with a question and I would like some help with it. I am currently sudying the A Level Mathematics for Edexcel book by Peter Hind in chapter 14 (differentiation), exercise 14.3. This is the question:


    A symmetrical cone is inside a sphere of radius r cm, with the vertex of the cone and the rim of the circular base in contact with the surface of the sphere. The centre of the base of the cone is x cm below the centre of the sphere.

    Show that the volume, V, of the cone is V = (π/3)(r-x)((r+x)^2)

    For the volume of the cone to be a maximum show that the value of x is r/3.

    Hence, or otherwise, find the maximum voume of the cone in terms of π and r.



    I was able to show that the volume of the cone equalled the one stated by expanding the brackets and using Pythagoras's Theorem, but I have stuck on the second part of the question: show that the value of x is r/3 at the maximum of the volume. I have tried expanding it and taking the differentation of both x and r (in other words, I have done both dV/dr and dV/dx), but have become stuck. Can ayone help me?
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    (Original post by fablereader)
    I was able to show that the volume of the cone equalled the one stated by expanding the brackets and using Pythagoras's Theorem, but I have stuck on the second part of the question: show that the value of x is r/3 at the maximum of the volume. I have tried expanding it and taking the differentation of both x and r (in other words, I have done both dV/dr and dV/dx), but have become stuck. Can ayone help me?
    What do you get for \frac{\mathrm{d}V}{\mathrm{d}x}?
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    (Original post by Zacken)
    What do you get for \frac{\mathrm{d}V}{\mathrm{d}x}?
    I expanded the brackets and got (πr^2/3) - (2πrx/3) - (3πx^2/3), which equals (π/3)(r^2 - 2rx - 3x^2)
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    (Original post by fablereader)
    I expanded the brackets and got (πr^2/3) - (2πrx/3) - (3πx^2/3), which equals (π/3)(r^2 - 2rx - 3x^2)
    That's very much correct, good work!

    So you know \frac{\mathrm{d}V}{\mathrm{d}x} = \frac{\pi}{3}\left(r^2 - 2rx - 3x^2\right).

    You also know that a function V attains its maximum (or minimum, but it's a maximum here) when \frac{\mathrm{d}V}{\mathrm{d}x} = 0.

    How would you solve \frac{\pi}{3}\left(r^2 - 2rx - 3x^2\right) = 0? Remember that it's just a quadratic. I would factorise it at \frac{\pi}{3} (r+x)(r-3x) = 0.

    This obviously gets you two stationary points, so you'll need to justify which one is a maximum and which one is a minimum... cough investigate the sign of \frac{\mathrm{d}^2V}{\mathrm{d}x  ^2}, is it less than zero or more than zero? What does this tell you about the nature of the two stationary points.
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    (Original post by Zacken)
    That's very much correct, good work!

    So you know \frac{\mathrm{d}V}{\mathrm{d}x} = \frac{\pi}{3}\left(r^2 - 2rx - 3x^2\right).

    You also know that a function V attains its maximum (or minimum, but it's a maximum here) when \frac{\mathrm{d}V}{\mathrm{d}x} = 0.

    How would you solve \frac{\pi}{3}\left(r^2 - 2rx - 3x^2\right) = 0? Remember that it's just a quadratic. I would factorise it at \frac{\pi}{3} (r+x)(r-3x) = 0.

    This obviously gets you two stationary points, so you'll need to justify which one is a maximum and which one is a minimum... cough investigate the sign of \frac{\mathrm{d}^2V}{\mathrm{d}x  ^2}, is it less than zero or more than zero? What does this tell you about the nature of the two stationary points.
    Ah, factorizing it! I briefly thought about it, but I didn't get how to do it and abandoned that road. Thanks!

    x would thus be either -r or r/3. Differentiating the differentiation, I get d^2V/dx^2 = (-2πr/3) - 2πx. If x = -r, I get -2πr/3 + 2πr. If x = r/3, I get -4πr/3. r is positive because it is the radius, and it cannot be negative as a result. Thus the second value has to be negative, and the first value has to be positive. Thus the second value of x (r/3) is a maximum, and the first value of x (-r) is a minimum.
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    (Original post by fablereader)
    Ah, factorizing it! I briefly thought about it, but I didn't get how to do it and abandoned that road. Thanks!
    If you aren't able to factorise, there's always the quadratic formula in a last ditch attempt (but a guaranteed one):

    \displaystyle

\begin{equation*} x = \frac{2r \pm \sqrt{4r^2 - 4(-3)(r^2)}}{-3 \times 2} = \frac{2r \pm \sqrt{16r^2}}{-6} = \frac{2r \pm 4r}{-6}\end{equation*}

    x would thus be either -r or r/3. Differentiating the differentiation, I get d^2V/dx^2 = (-2πr/3) - 2πx. If x = -r, I get -2πr/3 + 2πr. If x = r/3, I get -4πr/3. r is positive because it is the radius, and it cannot be negative as a result. Thus the second value has to be negative, and the first value has to be positive. Thus the second value of x (r/3) is a maximum, and the first value of x (-r) is a minimum.
    Perfect. Well done.
 
 
 
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