# Math Help - how to find maximum volume of cube inside cone

1. ## how to find maximum volume of cube inside cone

I am finding this one tricky.

Find the maximum volume of cube inscribed a right cone with radius 1, and height 3.

Why can't I do 3/1 = 3-x/x

If this was a cylinder, I'd have 3/1 = 3-h/r, and with a cube the sides are equal.

2. Originally Posted by Gul
I am finding this one tricky.

Find the maximum volume of cube inscribed a right cone with radius 1, and height 3.

Why can't I do 3/1 = 3-x/x

If this was a cylinder, I'd have 3/1 = 3-h/r, and with a cube the sides are equal.
Assuming you have defined x to be the sidelength of the cube then $\frac{3}{1} = \frac{3 - x}{\frac{x}{2}} = \frac{2(3 - x)}{x}$.

3. why is it a 1/2 - x/2, and not just x/2?

4. Originally Posted by Gul
why is it a 1/2 - x/2, and not just x/2?
You're right. I mixed up my triangles and made a typo (too busy being funny to be careful). It's fixed.

5. They get this:

3-X / 3 = Xsqrt(2) /2

6. Originally Posted by Gul
They get this:

3-X / 3 = Xsqrt(2) /2
Oboy. It's my day for dumb mistakes today. I have no tme now but I'll show where that comes from later.

Edit: The key is to note that when the cube is fit inside a cone, the longest side
is the diagonal of its top face. Read: