# Quadratic equation help: volume V = (1/3)(pi)(h)(R^2+Rr+r^2) of frustum of cone

#### Ryan493

##### New member
The formula for the volume of a frustum of a cone, V, is given below:

. . . . .$$\displaystyle V\, =\, \dfrac{1}{3}\, \pi\, h\, \left(R^2\, +\, Rr\, +\, r^2\right)$$

...where R is the major radius, r is the minor radius, and h is the height.

(i) Develop a quadratic equation representing the solution in terms of r.

(ii) Using the Quadratic Formula, solve for r, giving the answer to 2 decimal places.

(iii) Describe how you could also graphically solve this quadratic equation

Could anyone help im really lost

Thanks

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#### MarkFL

##### Super Moderator
Staff member
Hello, and welcome to FMH!

To express the given formula as a quadratic in $$r$$ in standard form, you first need to multiply both sides by $$\displaystyle \frac{3}{\pi h}$$...then what do you have?

#### Ryan493

##### New member
. . . . .$$\displaystyle \sqrt{\vphantom{\dfrac{3^2}{\pi}} \dfrac{3R\, -\, Rf\, -\, r}{\pi h}\,}$$

Am I on the right path

Thanks for your help

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#### MarkFL

##### Super Moderator
Staff member
What I had in mind would give you:

$$\displaystyle \frac{3V}{\pi h}=R^2+Rr+r^2$$

And then arrange in standard form:

$$\displaystyle r^2+Rr+R^2-\frac{3V}{\pi h}=0$$

Now, for part (ii) we are asked to use the quadratic formula to solve for $$r$$...can you proceed?

#### topsquark

##### Full Member
View attachment 11492

Am I on the right path

Thanks for your help
I assume you are trying to say
$$\displaystyle r = \sqrt{\dfrac{3R - Rr - r}{\pi h}}$$

I'm not even going to try to find out how you got this because there is a serious error right off. Notice that you have an "r" on both sides of your equation! So you know that can't possibly be the solution.

-Dan

#### Ryan493

##### New member
I assume you are trying to say
$$\displaystyle r = \sqrt{\dfrac{3R - Rr - r}{\pi h}}$$

I'm not even going to try to find out how you got this because there is a serious error right off. Notice that you have an "r" on both sides of your equation! So you know that can't possibly be the solution.

-Dan
I know Dan I have been searching to see how to even begin to do it. Do you have any advice representing the solution in terms of r. I'm so lost and appreciate your advice.

Thanks
Ryan

#### topsquark

##### Full Member
Similar with MarkFL's suggestion:
$$\displaystyle \pi h r^2 + \pi h Rr + \pi h R^2 - 3V = 0$$

(I'm simply avoiding fractions. Multiply both sides of MarkFL's equation by $$\displaystyle \pi h$$ to get my form if you can't see how to do it otherwise.)

This is a quadratic equation in r:
$$\displaystyle ( \pi h ) r^2 + ( \pi h R ) r + \left ( \pi h R^2 - 3V \right ) = 0$$
where $$\displaystyle a = \pi h$$, $$\displaystyle b = \pi h R$$, and $$\displaystyle c = \pi h R^2 - 3V$$

See what you can do with it from here.

Some general hints here.

1) You were asked in the first part to find a quadratic in r. That means you are looking for an expression $$\displaystyle ar^2 + br + c$$. A quadratic expression will always look like this.

2) You are asked to solve a quadratic equation. This is always of the form $$\displaystyle ax^2 + bx + c = 0$$ and if you can't figure out how to do it otherwise it will always have the solution set $$\displaystyle x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

When you have question about quadratic equations you can always always always find a starting point with these.

-Dan

#### Denis

##### Senior Member
Perhaps it would be easier if you let k = pi * h / 3
Then equation becomes: V = k(R^2 + Rr + r^2)