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# Stuck on differentiation question watch

1. 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?
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 ?
3. (Original post by Zacken)
What do you get for ?
I expanded the brackets and got (πr^2/3) - (2πrx/3) - (3πx^2/3), which equals (π/3)(r^2 - 2rx - 3x^2)
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 .

You also know that a function attains its maximum (or minimum, but it's a maximum here) when .

How would you solve ? Remember that it's just a quadratic. I would factorise it at .

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 , is it less than zero or more than zero? What does this tell you about the nature of the two stationary points.
5. (Original post by Zacken)
That's very much correct, good work!

So you know .

You also know that a function attains its maximum (or minimum, but it's a maximum here) when .

How would you solve ? Remember that it's just a quadratic. I would factorise it at .

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 , 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.
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):

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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