# Solving the Volume of an Object with Polynomial Functions

• April 23rd 2010, 01:13 PM
Xeaxy
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
$V(x) = -x^3 - x^2 + 6x$
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.
• April 23rd 2010, 01:55 PM
pickslides
Quote:

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

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

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 $-x^3-x^2+6x$ ??

Quote:

Originally Posted by Xeaxy
a. Factor the polynomial to find linear expressions for the height and the width.

Firstly take out $x$ as a common factor, what are you left with?
• April 23rd 2010, 02:06 PM
Xeaxy
Quote:

Originally Posted by pickslides
Are you sure the function is not $-x^3-x^2+6x$ ??

Oops, you're right. I fixed it.

Quote:

Originally Posted by pickslides
Firstly take out $x$ as a common factor, what are you left with?

$-x(x^2+x-6)$

I think.

EDIT: I factored completely and it comes out to $-x(x-2)(x+3)$. But I'm still confused as to how that represents the depth, height, and width of an object.
• April 23rd 2010, 02:32 PM
pickslides
Quote:

Originally Posted by Xeaxy
EDIT: I factored completely and it comes out to $-x(x-2)(x+3)$. 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 $-x$

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

so $-x^3-x^2+6x = x(2-x)(x+3) ,x<2$
• April 23rd 2010, 02:38 PM
Xeaxy
Quote:

Originally Posted by pickslides
so $-x^3-x^2+6x = x(2-x)(x+3) ,x<2$

Ok. So is $x<2$ the domain?

Also, what do the x-intercepts represent geometrically?
• April 23rd 2010, 02:54 PM
pickslides
Quote:

Originally Posted by Xeaxy
Ok. So is $x<2$ the domain?

Not exactly $x\in (0,2)$

Quote:

Originally Posted by Xeaxy
Ok. So is $x<2$
Also, what do the x-intercepts represent geometrically?

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

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

These $2$ intercepts show where $V(x)=0$ we want to consider everything in between.
• April 23rd 2010, 02:59 PM
Xeaxy
Quote:

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

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

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

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

Ok, everything makes a lot more sense now. Thanks so much!