# Thread: Solving the Volume of an Object with Polynomial Functions

1. ## Solving the Volume of an Object with Polynomial Functions

I've been stuck on a fairly difficult (well, for me at least) problem all day. It's concerning the volume in cubic feet of a CD holder:
The volume in cubic feet of a CD holder can be expressed as
$\displaystyle V(x) = -x^3 - x^2 + 6x$
or, when factored, as the product of its three dimensions, depth, height, and width. The depth is expressed as (2 - x). Assume the height is greater than the width.
The following are problems relating to the info above:
a. Factor the polynomial to find linear expressions for the height and the width.

b. Sketch a graph of the function. Find the x-intercepts. Explain what the x-intercepts represent geometrically.

c. Describe a realistic domain for the function V(x) and explain why you consider it to be realistic.

d. Find the maximum volume of the CD holder.
I've been doing a lot of problems like this in my Algebra 2 class, and I just need a simple explanation on what to do. I know how to graph a polynomial function like this, so (b.) shouldn't be a problem.

2. Originally Posted by Xeaxy
The volume in cubic feet of a CD holder can be expressed as

$\displaystyle V(x) = -x^2 - x^2 + 6x$

or, when factored, as the product of its three dimensions, depth, height, and width. The depth is expressed as (2 - x). Assume the height is greater than the width.
Are you sure the function is not $\displaystyle -x^3-x^2+6x$ ??

Originally Posted by Xeaxy
a. Factor the polynomial to find linear expressions for the height and the width.
Firstly take out $\displaystyle x$ as a common factor, what are you left with?

3. Originally Posted by pickslides
Are you sure the function is not $\displaystyle -x^3-x^2+6x$ ??
Oops, you're right. I fixed it.

Originally Posted by pickslides
Firstly take out $\displaystyle x$ as a common factor, what are you left with?
$\displaystyle -x(x^2+x-6)$

I think.

EDIT: I factored completely and it comes out to $\displaystyle -x(x-2)(x+3)$. But I'm still confused as to how that represents the depth, height, and width of an object.

4. Originally Posted by Xeaxy
EDIT: I factored completely and it comes out to $\displaystyle -x(x-2)(x+3)$. But I'm still confused as to how that represents the depth, height, and width of an object.
It does look confusing because you have what looks like a negative dimension in $\displaystyle -x$

But they have given you a hint above in $\displaystyle d =(2-x)$

so $\displaystyle -x^3-x^2+6x = x(2-x)(x+3) ,x<2$

5. Originally Posted by pickslides
so $\displaystyle -x^3-x^2+6x = x(2-x)(x+3) ,x<2$
Ok. So is $\displaystyle x<2$ the domain?

Also, what do the x-intercepts represent geometrically?

6. Originally Posted by Xeaxy
Ok. So is $\displaystyle x<2$ the domain?
Not exactly $\displaystyle x\in (0,2)$

Originally Posted by Xeaxy
Ok. So is $\displaystyle x<2$
Also, what do the x-intercepts represent geometrically?
The other intercept $\displaystyle x=-3$ is no use to us, we don't want negative dimension.

Look at a graph of the function. $\displaystyle V(x) > 0,x \in (0,2)$

These $\displaystyle 2$ intercepts show where $\displaystyle V(x)=0$ we want to consider everything in between.

7. Originally Posted by pickslides
Not exactly $\displaystyle x\in (0,2)$

The other intercept $\displaystyle x=-3$ is no use to us, we don't want negative dimension.

Look at a graph of the function. $\displaystyle V(x) > 0,x \in (0,2)$

These $\displaystyle 2$ intercepts show where $\displaystyle V(x)=0$ we want to consider everything in between.
Ok, everything makes a lot more sense now. Thanks so much!