Optimization largest possible volume problem

pynergee
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Homework Statement


A right circular cylinder is inscribed in a cone with height h and base radius r. Find the largest possible volume of such a cylinder.

Volume of a cylinder = (pi)(r^2)h
Volume of a cone = (1/3)(pi)(r^2)h


Homework Equations


Volume of a cylinder = (pi)(r^2)h
Volume of a cone = (1/3)(pi)(r^2)h



The Attempt at a Solution


Technically this is my roommate's problem, he is in Calc1. He has been having some problems with this one, and I'm in Differential Equations, and I can't remember how to really do this one. I know you want to find the derivative of the cylinder, and find when it is equal to zero, but I am stumped on how to approach the problem
 
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solve for the height of the cylinder in terms of its radius then look at optimisation through substitution or lagrange multipliers
 
I know that, but how? Would you use similar triangles or something like that?
 
you could do that...

draw a vertical slice of the cone thorough its centre. The cone appears as an isoceles triangle, whilst the cylinder is a rectangle inscribed in the triangle. Use some trig to derive the relation between cylinder height & base
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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