# Show using the intermediate value theorem

1. Sep 30, 2009

### John O' Meara

Use the Intermediate Value Theorem to show that there is a cylinder of height h and radius less than r whose volume is equal to that of a cone of height h and radius r.
IVT states that: if f is continuous on a closed interval [a,b] and k is any number between f(a) and f(b), inclusive, then there is at least one number x in the interval [a,b] such that f(x)=k.
The volume of the cylinder is $$\pi x^2h$$ and the volume of the cone is $$\frac{\pi}{3} r^2h$$, where x< r. If r > x then, the curve $$\frac{\pi}{3} r^2h$$ is streched more than the curve $$\pi x^2h$$, i.e it is closer to the y-axis than the curve [\pi x^2h[/tex], so I cannot find k. And is not a<0 and b>r? Then f(a)<0 and f(b)>f(r)?I am studing this on my own. Please help. Thanks.

Last edited: Sep 30, 2009
2. Sep 30, 2009

### boboYO

I dont see how y=x^2/3 is closer to the y axis than y=x^2? It's the other way around

and if you haven't solved the problem,
look at the interval [0,r]

3. Sep 30, 2009

### HallsofIvy

Staff Emeritus
As you say, a cone of height h and radius r has volume $\frac{1}{3}\pi r^2 h$.

What is the volume of a cylinder of height h and radius 0? What is the volume of a cylinder of height h and radius r? Now, what does the intermediate value theorem tell you?

4. Oct 1, 2009

### John O' Meara

The volume of the cylinder is 0 and $$\pi r^2h$$ respectively. I think you mean that f(a)=0 and $$f(b)= \pi r^2h$$. In that case surely there is an infinite number of values that k can have between f(a) and f(b). But we only want one value of k such that the volume of the cylinder equals the volume of the cone. I just don't understand yet.

5. Oct 1, 2009

### HallsofIvy

Staff Emeritus
Yes, there are an infinite number of values between 0 and $\pi h r^2$.

And one of them is $(1/3)\pi h r^2$!

The intermediate value theorem tells you that as $\rho$ goes from 0 to r $\pi h \rho^2$ takes on all values between 0 and $\pi h r^2$.